# For each series, find an explicit formula for the sequence of partial sums and determine if the series converges. sum_{n=1}^inftyfrac{1}{n(n+1)}

illusiia 2021-02-18 Answered
For each series, find an explicit formula for the sequence of partial sums and determine if the series converges.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}$
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## Expert Answer

joshyoung05M
Answered 2021-02-19 Author has 97 answers
Given: The series $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}$
To determine: The explicit formula for the sequence of partial sum and convergence of the series.
Explanation:
The general term (k-th term) of the sequence of partial sum is of a series is the sum of first k terms.
Here the given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}$
So general term of sequence of partial sum of this series is $\sum _{n=1}^{n=k}\frac{1}{n\left(n+1\right)}$
This general can be re-written as
${t}_{k}=\sum _{n=1}^{n=k}\frac{1}{n\left(n+1\right)}$
$⇒{t}_{k}=\sum _{n=1}^{n=k}\left(\frac{1}{n}-\frac{1}{n+1}\right)$
$⇒{t}_{k}=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{k}-\frac{1}{k+1}\right)$
$⇒{t}_{k}=\left(1-\frac{1}{k+1}\right)$
Hence general term of sequence of partial sum of this series is ${t}_{k}=\left(1-\frac{1}{k+1}\right)$
So sequence of partial sum is ${\left\{1-\frac{1}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}$
Now it is known that $\sum _{n=1}^{\mathrm{\infty }}{a}_{n}=\underset{k\to \mathrm{\infty }}{lim}\sum _{n=1}^{k}{a}_{n}$
Therefore here
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}=\underset{k\to \mathrm{\infty }}{lim}\sum _{n=1}^{k}\frac{1}{n\left(n+1\right)}$
$⇒\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}=\underset{k\to \mathrm{\infty }}{lim}\left(1-\frac{1}{k+1}\right)$
$⇒\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n\left(n+1\right)}=1$
Hence the series is convergent as converges to 1.
Answer:
The sequence of partial sum is ${\left\{1-\frac{1}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}$
And the series is convergent as converges to 1.
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Jeffrey Jordon
Answered 2021-12-16 Author has 2070 answers

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