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? ∑k=1∞(2k−2k+1)

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?  ∑k=1∞(2k−2k+1)

escumantsu · Accepted Answer

To determine the convergence/divergence of the series.  Given:  ∑k=1∞(2k−2k+1)  a) Simplify as,  ∑k=1∞(2k−2k+1)=2∑k=1∞(1k−1k+1)  Sn=2∑k=1n(1k−1k+1)=(11−11+1)+(12−12+1)+...+(1n−1n+1)  Sn=2(1−1n+1)  limn→∞Sn=2limn→∞(1−1n+1)  =2(1−0)  =2  ⇒∑k=1∞(2k−2k+1)=2  ⇒∑k=1∞(2k−2k+1)→2  Hence, the series converges to 2.  (b) The sum of the first 5 term of the sequence is,  ∑k=15(2k−2k+1)=(21−21+1)+(22−22+1)+(23−23+1)+(24−24+1)+(25−25+1)  =2−26  =53  (c) The general term for the terms of the partial sum is,  Sn=2∑k=1n(1k−1k+1)=(11−11+1)+(12−12+1)+...+(1n−1n+1)  Sn=2(1−1n+1)

Jeffrey Jordon · Accepted Answer

Answer is given below (on video)

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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New Questions in Calculus 2