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

Question

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.

Answer 1

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.

Answer 2

Hint:

$$ 2^n > n^3$$

for $n\geq 10$.

Answer 3

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

Answer 4

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 $S_N\to\color{red}{3}$.

Answer 5

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 $n > n_0$.

Answer 6

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