Question

A) Does this series converge? If yes, towards what number? B) Find the first 5 terms of the sequence of partial sums in this series. C) What is the general term of this sequence of partial sums? sum_{k=1}^infty(frac{2}{k}-frac{2}{k+1})

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ANSWERED
asked 2021-01-13
A) Does this series converge? If yes, towards what number?
B) Find the first 5 terms of the sequence of partial sums in this series.
C) What is the general term of this sequence of partial sums?
\(\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})\)

Answers (1)

2021-01-14
To determine the convergence/divergence of the series.
Given:
\(\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})\)
a) Simplify as,
\(\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})=2\sum_{k=1}^\infty(\frac{1}{k}-\frac{1}{k+1})\)
\(S_n=2\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+1})=(\frac{1}{1}-\frac{1}{1+1})+(\frac{1}{2}-\frac{1}{2+1})+...+(\frac{1}{n}-\frac{1}{n+1})\)
\(S_n=2(1-\frac{1}{n+1})\)
\(\lim_{n\to\infty}S_n=2\lim_{n\to\infty}(1-\frac{1}{n+1})\)
\(=2(1-0)\)
\(=2\)
\(\Rightarrow\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})=2\)
\(\Rightarrow\sum_{k=1}^\infty(\frac{2}{k}-\frac{2}{k+1})\to2\)
Hence, the series converges to 2.
(b) The sum of the first 5 term of the sequence is,
\(\sum_{k=1}^5(\frac{2}{k}-\frac{2}{k+1})=(\frac{2}{1}-\frac{2}{1+1})+(\frac{2}{2}-\frac{2}{2+1})+(\frac{2}{3}-\frac{2}{3+1})+(\frac{2}{4}-\frac{2}{4+1})+(\frac{2}{5}-\frac{2}{5+1})\)
\(=2-\frac{2}{6}\)
\(=\frac{5}{3}\)
(c) The general term for the terms of the partial sum is,
\(S_n=2\sum_{k=1}^n(\frac{1}{k}-\frac{1}{k+1})=(\frac{1}{1}-\frac{1}{1+1})+(\frac{1}{2}-\frac{1}{2+1})+...+(\frac{1}{n}-\frac{1}{n+1})\)
\(S_n=2(1-\frac{1}{n+1})\)
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