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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.

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    $\begingroup$ 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. $\endgroup$
    – sciona
    Jan 22, 2015 at 18:03
  • $\begingroup$ The ratio test suffices. $\endgroup$
    – MJD
    Jan 22, 2015 at 18:03
  • $\begingroup$ Oh okay.. well, I guess my book ( Calculus by Apostol ) is yet to introduce the Rabbe's Test or the ratio test. So, ... $\endgroup$
    – MathMan
    Jan 22, 2015 at 18:04
  • $\begingroup$ Display environments and display style are not appropriate for titles, please see meta. $\endgroup$
    – dustin
    Jan 25, 2015 at 18:22

6 Answers 6

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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.

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Hint:

$$ 2^n > n^3$$

for $n\geq 10$.

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    $\begingroup$ Thank you for the hint! Got it :) $\endgroup$
    – MathMan
    Jan 22, 2015 at 21:59
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Ratio test works fine, and so does the root test.

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  • $\begingroup$ Thank you!! I worked it out :) $\endgroup$
    – MathMan
    Jan 22, 2015 at 22:03
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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 $S_N\to\color{red}{3}$.

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    $\begingroup$ Interesting!! Thank you for the answer :) $\endgroup$
    – MathMan
    Jan 22, 2015 at 22:01
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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 $n > n_0$.

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  • $\begingroup$ Thank you for the answer ! :) $\endgroup$
    – MathMan
    Jan 22, 2015 at 21:58
  • $\begingroup$ No problem, glad it helped. $\endgroup$
    – user203856
    Jan 22, 2015 at 22:02
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Hint: $$\lim_{n\to\infty}\Bigg|\frac{a_{n+1}}{a_n}\Bigg| =\lim_{n\to\infty} \frac{n+1}{2^{n+1}}\frac{2^n}{n } = \ldots$$

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  • $\begingroup$ You're welcome, I'm glad I could help. $\endgroup$ Jan 22, 2015 at 22:01

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