Question: "Evaluate:∑n=1∞n2+1n⋅2n−1"

Zaccagliaodi

Answered question

2022-02-26

Evaluate:

\sum_{n = 1}^{\infty} \frac{n^{2} + 1}{n \cdot 2^{n - 1}}

Capodarcod0f

Beginner2022-02-27Added 6 answers

What you can do is split it in two

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

\frac{1}{n 2^{n - 1}} = 2 \frac{1}{n 2^{n}} = 2 \int_{0}^{\frac{1}{2}} x^{k - 1} d x

Hence

\sum_{k = 1}^{N} \frac{1}{2^{k - 1}} = 2 \sum_{k = 1}^{N} \int_{0}^{\frac{1}{2}} x^{k - 1} d x

= 2 \int_{0}^{\frac{1}{2}} \sum_{k = 1}^{N} x^{k - 1} d x

= 2 \int_{0}^{\frac{1}{2}} \frac{1 - x^{N}}{1 - x} d x

From this, you have

\sum_{k = 1}^{N} \frac{1}{k 2^{k - 1}} = 2 \int_{0}^{\frac{1}{2}} \frac{d x}{1 - x} - 2 \int_{0}^{\frac{1}{2}} \frac{x^{N}}{1 - x} d x

Making

N \to + \infty

, I let you show that

\int_{0}^{\frac{1}{2}} \frac{x^{N}}{1 - x} d x \to 0

Hence

\sum_{k = 1}^{+ \infty} \frac{1}{k 2^{k - 1}} = 2 \int_{0}^{\frac{1}{2}} \frac{d x}{1 - x} = \log (4)

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