Telescoping Series

Section 7.2 Telescoping Series

Besides geometric series, there is another special type of series called a telescoping series. Many telescoping series also have a finite sum that we can compute exactly, because of a cancellation pattern.

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If a series has partial sums which are telescoping sums, then we can determine a closed-form formula for the partial sums, and evaluate the series using limits.

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Example 7.2.1. Sum of 1/n(n+1).

Consider the series,

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\begin{equation*} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} = \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots \end{equation*}

The term can be rewritten as two fractions, using partial fraction decomposition,

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\begin{equation*} \frac{1}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \end{equation*}

Solving for \(A\) and \(B\text{,}\) we get \(A = 1, B = -1\text{,}\) so,

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\begin{equation*} \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1} \end{equation*}

Then, consider the partial sums of the series, which are given by,

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\begin{align*} s_N \amp= \sum_{n=1}^N \brac{\frac{1}{n} - \frac{1}{n+1}}\\ \amp= \underbrace{\frac{1}{1} - \frac{1}{2}}_{n=1} + \underbrace{\frac{1}{2} - \frac{1}{3}}_{n=2} + \underbrace{\frac{1}{3} - \frac{1}{4}}_{n=3} + \dots + \underbrace{\frac{1}{N} - \frac{1}{N+1}}_{n=N} \end{align*}

Notice that the \(\frac{1}{2}\)s cancel out, the \(\frac{1}{3}\)s also cancel out, etc. In general, each negative term is cancelled out by the positive term 1 term to the right. Then, the only terms that are not cancelled out are the leading \(\frac{1}{1}\) and the ending \(-\frac{1}{N+1}\text{.}\) Therefore,

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\begin{equation*} s_N = 1 - \frac{1}{N+1} \end{equation*}

This is a formula which gives the partial sum of the series \(s_N\) for any \(N\text{.}\) In other words, this formula gives the sum of the first \(N\) terms of the series. For example,

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\begin{align*} s_1 \amp= 1 - \frac{1}{2} = \frac{1}{2}\\ s_2 \amp= 1 - \frac{1}{3} = \frac{2}{3}\\ s_3 \amp= 1 - \frac{1}{4} = \frac{3}{4}\\ s_{100} \amp= 1 - \frac{1}{101} = \frac{100}{101} \end{align*}

Recall that the sum of a series is the limit of its partial sums as \(N \to \infty\text{.}\) Notice that the partial sums \(s_N\) are approaching 1. In particular, as \(N \to \infty\text{,}\)

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\begin{align*} \sum_{n=1}^{\infty} \frac{1}{n(n+1)} \amp= \lim_{N \to \infty} s_N\\ \amp= \lim_{N \to \infty} \brac{1 - \underbrace{\frac{1}{N+1}}_{\to 0}}\\ \amp= 1 \end{align*}

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Remark 7.2.2.

Multiplying the above sum by 2 gives a famous series,

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\begin{equation*} 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \dots \end{equation*}

The sequence \(1, 3, 6, 10, 15, 21, \dots\) are the so-called triangular numbers, so this series is the sum of the reciprocals of the triangular numbers. From above, its sum is,

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\begin{align*} 1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} + \frac{1}{15} + \frac{1}{21} + \dots \amp= 2 \cdot \sum_{n=1}^{\infty} \frac{1}{n(n+1)}\\ \amp= 2 \cdot 1 = 2 \end{align*}

This sum was posed to Leibniz in 1672 by Christiaan Huygens, Dutch mathematician and physicist. He evaluated it by considering the series divided by 2, then wrote it as a telescoping sum. Also, Jacob Bernoulli studied this series around 1689, an determined the sum to be 2.

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In general, the idea is that if a series can be written as the difference between two terms that are similar, then most of them will cancel out, and we can find the sum of the series by finding the limit of the partial sums.

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Example 7.2.3. Sum of 1/(n^2-1).

Consider the series,

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\begin{align*} \sum_{n=2}^{\infty} \frac{1}{n^2-1} \amp= \frac{1}{2^2 - 1} + \frac{1}{3^2 - 1} + \frac{1}{4^2 - 1} + \frac{1}{5^2 - 1} + \dots\\ \amp= \frac{1}{3} + \frac{1}{8} + \frac{1}{15} + \frac{1}{24} + \dots \end{align*}

We can also use partial fraction decomposition to split the term into two fractions, by factoring the denominator.

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\begin{equation*} \frac{1}{(n-1)(n+1)} = \frac{A}{n - 1} + \frac{B}{n + 1} \end{equation*}

Solving for \(A\) and \(B\text{,}\) we get \(A = \frac{1}{2}, B = -\frac{1}{2}\text{,}\) so,

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\begin{equation*} \frac{1}{n^2 - 1} = \frac{1/2}{n-1} + \frac{-1/2}{n+1} \end{equation*}

We can factor out the \(\frac{1}{2}\) for simplicity, and focus on the part involving \(n\text{,}\)

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\begin{equation*} \frac{1}{n^2-1} = \frac{1}{2} \brac{\frac{1}{n-1} - \frac{1}{n+1}} \end{equation*}

Then, the partial sums are given by,

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\begin{equation*} s_N = \sum_{n=2}^N \frac{1}{n^2 - 1} = \frac{1}{2} \sum_{n=2}^N \brac{\frac{1}{n-1} - \frac{1}{n+1}} \end{equation*}

Then, expanding,

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\begin{equation*} s_N = \frac{1}{2} \brac{\underbrace{\frac{1}{1} - \frac{1}{3}}_{n=2} + \underbrace{\frac{1}{2} - \frac{1}{4}}_{n=3} + \underbrace{\frac{1}{3} - \frac{1}{5}}_{n=4} + \dots + \underbrace{\frac{1}{N-3} - \frac{1}{N-1}}_{n=N-2} + \underbrace{\frac{1}{N-2} - \frac{1}{N}}_{n=N-1} + \underbrace{\frac{1}{N-1} - \frac{1}{N+1}}_{n=N}} \end{equation*}

Notice that the \(\frac{1}{3}\)s cancel, the \(\frac{1}{4}\)s also cancel, etc. In general, each negative term is cancelled out by the positive term 3 terms to the right. Then, the only terms left are the leading \(\frac{1}{1}, \frac{1}{2}\text{,}\) and the ending \(-\frac{1}{N}\) and \(-\frac{1}{N+1}\text{.}\) Then,

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\begin{equation*} s_N = \frac{1}{2} \brac{\frac{3}{2} - \frac{1}{N} - \frac{1}{N+1}} \end{equation*}

Then, as \(N \to \infty\text{,}\)

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\begin{align*} \sum_{n=2}^{\infty} \frac{1}{n^2 - 1} \amp= \lim_{N \to \infty} s_N\\ \amp= \lim_{N \to \infty} \frac{1}{2} \brac{\frac{3}{2} - \underbrace{\frac{1}{N}}_{\to 0} - \underbrace{\frac{1}{N+1}}_{\to 0}}\\ \amp= \frac{1}{2} \cdot \frac{3}{2} = \frac{3}{4} \end{align*}

Therefore, the series converges to \(\frac{3}{4}\text{.}\)

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Remark 7.2.4.

The Bernoulli brothers found that,

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\begin{equation*} \boxed{\sum_{n=2}^{\infty} \frac{1}{n^2 - 1} = \frac{3}{4}} \end{equation*}

However, they couldn’t solve \(\sum_{n=1}^{\infty} \frac{1}{n^2}\) (a similar series, but without \(-1\) in the denominator), which is much more difficult to evaluate.

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Example 7.2.5. Telescoping with Gap 2.

Consider the series,

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\begin{equation*} \sum_{n=2}^{\infty} \brac{\frac{1}{n} - \frac{1}{n+2}} \end{equation*}

Notice the series starts at \(n=2\text{,}\) not \(n=1\text{.}\) The partial sums are,

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\begin{align*} s_N \amp= \sum_{n=2}^N \brac{\frac{1}{n} - \frac{1}{n+2}}\\ \amp= \underbrace{\frac{1}{2} - \frac{1}{4}}_{n=2} + \underbrace{\frac{1}{3} - \frac{1}{5}}_{n=3} + \underbrace{\frac{1}{4} - \frac{1}{6}}_{n=4} + \dots + \brac{\frac{1}{N - 2} - \frac{1}{N}} + \brac{\frac{1}{N - 1} - \frac{1}{N + 1}} + \brac{\frac{1}{N} - \frac{1}{N+2}} \end{align*}

The \(\frac{1}{4}\)s cancel out, \(\frac{1}{5}\)s also, etc. In general, each negative term cancels out with the positive term 3 terms to the right. In the end, only the \(\frac{1}{2}\) and \(\frac{1}{3}\) are left, along with \(-\frac{1}{N + 1}\) and \(-\frac{1}{N + 2}\text{.}\) Then,

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\begin{equation*} s_N = \frac{5}{6} - \frac{1}{N + 1} - \frac{1}{N + 2} \end{equation*}

Then, as \(N \to \infty\text{,}\)

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\begin{align*} \sum_{n=2}^{\infty} \brac{\frac{1}{n} - \frac{1}{n+2}} \amp= \lim_{N \to \infty} s_N\\ \amp= \lim_{N \to \infty} \brac{\frac{5}{6} - \underbrace{\frac{1}{N + 1}}_{\to 0} - \underbrace{\frac{1}{N + 2}}_{\to 0}}\\ \amp= \frac{5}{6} \end{align*}

Therefore, the series converges to \(\frac{5}{6}\text{.}\)

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Telescoping series don’t always converge, as in the next example.

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Example 7.2.6. A Divergent Telescoping Series.

Consider the series,

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\begin{equation*} \sum_{n=1}^{\infty} \ln{\brac{\frac{n}{n + 1}}} \end{equation*}

We can’t use partial fraction decomposition here, but instead we use the logarithm property \(\ln{\brac{\frac{a}{b}}} = \ln{a} - \ln{b}\text{,}\) to rewrite it as,

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\begin{equation*} \ln{\brac{\frac{n}{n + 1}}} = \ln{n} - \ln{(n+1)} \end{equation*}

Notice again that we have the difference between 2 terms that are similar. The partial sums are,

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\begin{align*} s_N \amp= \sum_{n=1}^N \brac{\ln{n} - \ln{(n+1)}}\\ \amp= \underbrace{\ln{1} - \ln{2}}_{n=1} + \underbrace{\ln{2} - \ln{3}}_{n=2} + \underbrace{\ln{3} - \ln{4}} + \dots + \underbrace{\ln{N} - \ln{(N+1)}}_{n=N} \end{align*}

Each of the negative terms is cancelled out by the positive term 1 term to the right. Then, the only terms that are not cancelled out are the leading \(\ln{1}\) (recall \(\ln{1} = 0\)) and the ending \(-\ln{(N+1)}\text{.}\) Therefore,

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\begin{equation*} s_N = -\ln{(N+1)} \end{equation*}

Then, as \(N \to \infty\text{,}\)

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\begin{align*} \sum_{n=1}^{\infty} \ln{\brac{\frac{n}{n + 1}}} \amp= \lim_{N \to \infty} s_N\\ \amp= \lim_{N \to \infty} -\underbrace{\ln{(N+1)}}_{\to \infty}\\ \amp= -\infty \end{align*}

Therefore, the series diverges to \(-\infty\text{.}\)

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Subsection 7.2.1 Summary of Telescoping Series

Rewrite the term as a difference of 2 terms (if it isn’t already), often using partial fraction decomposition.
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Write out the first few terms (like \(n=1, 2, 3\)), and the last few terms (\(n=N-1, N\)) to see the cancellation pattern.
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Observe the cancellation pattern: “Each positive term cancels out with the negative term 1/2/3/4 to its right”. Cancel out the terms. Typically, you’ll be left with a few at the start, and a few at the end.
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Take the limit of the simplified partial sum as \(N \to \infty\text{.}\)
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You can write out as many terms as you need to recognize the pattern. If the pattern is not immediately obvious, then try writing out more terms.

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To recognize that a series is telescoping, the 2 main things to look for are:

Are two similar terms being subtracted?
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Or, can you do partial fraction decomposition?
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Subsection 7.2.2 Examples

Exercise Group 7.2.1. Telescoping Series (★).

Find the sum of each telescoping series, or show that it diverges.

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(a)

\(\displaystyle\sum_{n=1}^{\infty}\brac{\frac{1}{n+2}-\frac{1}{n+3}}\)

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Hint.

\(s_N = \frac{1}{3} - \frac{1}{N+3}\)

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Answer.

\(\frac{1}{3}\)

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(b)

\(\displaystyle\sum_{n=1}^{\infty}\brac{\frac{1}{n+1}-\frac{1}{n+2}}\)

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Hint.

\(s_N = \frac{1}{2} - \frac{1}{N+2}\)

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Answer.

\(\frac{1}{2}\)

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(c)

\(\displaystyle\sum_{n=4}^{\infty}\brac{\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}}\)

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Hint.

\(s_N = \frac{1}{2} - \frac{1}{\sqrt{N+1}}\)

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Answer.

\(\frac{1}{2}\)

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(d)

\(\displaystyle\sum_{n=1}^{\infty}\brac{\frac{3}{n^2}-\frac{3}{(n+1)^2}}\)

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Hint.

\(s_N = 3 - \frac{3}{(N+1)^2}\)

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Answer.

\(3\)

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(e)

\(\displaystyle\sum_{n=1}^{\infty}\brac{\sqrt{n+1}-\sqrt{n}}\)

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Hint.

\(s_N = \sqrt{N+1} - 1\)

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Answer.

Diverges to \(\infty\)

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(f)

\(\displaystyle\sum_{n=1}^{\infty}\brac{\frac{1}{n^8}-\frac{1}{(n+1)^8}}\)

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Hint.

\(s_N = 1 - \frac{1}{(N+1)^8}\)

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Answer.

\(1\)

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(g)

\(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+3)(n+5)}\)

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Hint.

Partial fractions, \(\frac{1}{(n+3)(n+5)} = \frac{1}{2}\brac{\frac{1}{n+3} - \frac{1}{n+5}}\text{.}\) \(s_N = \frac{1}{2}\brac{\frac{9}{20} - \frac{1}{N+4} - \frac{1}{N+5}}\)

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Answer.

\(\frac{9}{40}\)

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(h)

\(\displaystyle\sum_{n=1}^{\infty}\frac{9}{n(n+3)}\)

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Hint.

Partial fractions, \(\frac{9}{n(n+3)} = \frac{3}{n} - \frac{3}{n+3}\text{.}\) \(s_N = \frac{11}{2} - \frac{3}{N+1} - \frac{3}{N+2} - \frac{3}{N+3}\)

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Answer.

\(\frac{11}{2}\)

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(i)

\(\displaystyle\sum_{n=1}^{\infty}\frac{4}{n(n+2)}\)

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Hint.

Partial fractions, \(\frac{4}{n(n+2)} = \frac{2}{n} - \frac{2}{n+2}\text{.}\) \(s_N = 3 - \frac{2}{N+1} - \frac{2}{N+2}\)

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Answer.

\(3\)

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(j)

\(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+3)}\)

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Hint.

Partial fractions, \(\frac{1}{(2n+1)(2n+3)} = \frac{1}{2}\brac{\frac{1}{2n+1} - \frac{1}{2n+3}}\text{.}\) \(s_N = \frac{1}{6} - \frac{1}{2(2N+3)}\)

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Answer.

\(\frac{1}{6}\)

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(k)

\(\displaystyle\sum_{n=1}^{\infty}\frac{1}{(n+1)(n+2)}\)

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Hint.

Partial fractions, \(\frac{1}{(n+1)(n+2)} = \frac{1}{n+1} - \frac{1}{n+2}\text{.}\) \(s_N = \frac{1}{2} - \frac{1}{N+2}\)

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Answer.

\(\frac{1}{2}\)

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(l)

\(\displaystyle\sum_{n=1}^{\infty}\frac{5}{n(n+1)}\)

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Hint.

Find the sum of each telescoping series, or show that it diverges.

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(a)

\(\displaystyle\sum_{n=3}^{\infty}\frac{4}{(4n-3)(4n+1)}\)

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Hint.

Partial fractions, \(\frac{4}{(4n-3)(4n+1)} = \frac{1}{4n-3} - \frac{1}{4n+1}\text{.}\) \(s_N = \frac{1}{9} - \frac{1}{4N+1}\)

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Answer.

\(\frac{1}{9}\)

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(b)

\(\displaystyle\sum_{n=0}^{\infty}\frac{1}{(3n+1)(3n+4)}\)

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Hint.

Partial fractions, \(\frac{1}{(3n+1)(3n+4)} = \frac{1}{3}\brac{\frac{1}{3n+1} - \frac{1}{3n+4}}\text{.}\) \(s_N = \frac{1}{3}\brac{1 - \frac{1}{3N+4}}\)

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Answer.

\(\frac{1}{3}\)

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