Evaluate: \sum_{n=1}^\infty\frac{n^2+1}{n\cdot2^{n-1}}

Zaccagliaodi

Zaccagliaodi

Answered question

2022-02-26

Evaluate:
n=1n2+1n2n1

Answer & Explanation

Capodarcod0f

Capodarcod0f

Beginner2022-02-27Added 6 answers

What you can do is split it in two
n=1n2+1n2n1=n=1+n(12)n1+n=1+(12)n1n
1n2n1=21n2n=2012xk1dx
Hence
k=1N12k1=2k=1N012xk1dx
=2012k=1Nxk1dx
=20121xN1xdx
From this, you have
k=1N1k2k1=2012dx1x2012xN1xdx
Making N+, I let you show that
012xN1xdx0
Hence
k=1+1k2k1=2012dx1x=log(4)

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?