# Convergence of $\sum\limits_{n=1}^\infty \frac {n+1}{2^n}$.

Test Convergence of $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n}$$

Attempt: $$\sum\limits_{n=1}^\infty \dfrac {n+1}{2^n} = \sum\limits_{n=1}^\infty \dfrac {n }{2^n} + \sum\limits_{n=1}^\infty \dfrac {1}{2^n}$$

The second summation is definitely convergent. So, we need to just investigate if the first summation is convergent or not.

Let $$X = \sum\limits_{n=1}^\infty \dfrac {n }{2^n}$$

Is there a way to test convergence of this summation without the integral test?

Thank you very much for your help.

• Using $n \le 2^{n/2}$ for $n \ge 4$ and applying comparison test? unless you identify it as an AGP and evaluate it explicitly. – sciona Jan 22 '15 at 18:03
• The ratio test suffices. – MJD Jan 22 '15 at 18:03
• Oh okay.. well, I guess my book ( Calculus by Apostol ) is yet to introduce the Rabbe's Test or the ratio test. So, ... – MathMan Jan 22 '15 at 18:04
• Display environments and display style are not appropriate for titles, please see meta. – dustin Jan 25 '15 at 18:22

Try the comparison test with $\Sigma \frac {1}{n^2}$ , i.e., show that for a fixed index your general term $\frac {n}{2^n} < \frac {1}{n^2}$ * and then use that $\frac {n}{2^n}$ decreases faster than $\frac {1}{n^2}$.

EDIT *By this I mean that there is a value $n_0$ so that $\frac {n}{2^n} < \frac {1}{n^2}$ is true for all $n > n_0$.

• Thank you for the answer ! :) – MathMan Jan 22 '15 at 21:58
• No problem, glad it helped. – user203856 Jan 22 '15 at 22:02

I think that D'alambert's test (about the quotient) will be good here.

Explanation:

We need to check if $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1$ (then it converges). In our case,

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{\frac{n+1}{2\cdot 2^n}}{\frac{n}{2^n}}=\lim_{n\to\infty}\frac{n+1}{2n}=\frac{1}{2}$$

Because the limit is smaller than 1, the sum converges.

Hint:

$$2^n > n^3$$

for $n\geq 10$.

• Thank you for the hint! Got it :) – MathMan Jan 22 '15 at 21:59

Ratio test works fine, and so does the root test.

• Thank you!! I worked it out :) – MathMan Jan 22 '15 at 22:03

Let: $$S_N = \sum_{n=1}^{N}\frac{n+1}{2^n}.$$ We have: $$\begin{eqnarray*} \frac{S_N}{2}&=&S_N-\frac{S_N}{2}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=1}^{N}\frac{n+1}{2^{n+1}}=\sum_{n=1}^{N}\frac{n+1}{2^n}-\sum_{n=2}^{N+1}\frac{n}{2^{n}}\\&=&1+\sum_{n=2}^{N}\frac{1}{2^n}-\frac{N+1}{2^{N+1}}=\frac{3}{2}-\frac{N+3}{2^{N+1}}.\end{eqnarray*}$$ Since $\frac{N+2}{2^{N+1}}\to 0$ as $N\to +\infty$, we have $S_N\to\color{red}{3}$.

• Interesting!! Thank you for the answer :) – MathMan Jan 22 '15 at 22:01

Hint: $$\lim_{n\to\infty}\Bigg|\frac{a_{n+1}}{a_n}\Bigg| =\lim_{n\to\infty} \frac{n+1}{2^{n+1}}\frac{2^n}{n } = \ldots$$

• You're welcome, I'm glad I could help. – Aaron Maroja Jan 22 '15 at 22:01