3

What would be the simplest way to prove that $\sum\limits_{i=1}^{\infty}\dfrac{i}{2^i}$ converges?

Martin Sleziak
  • 56,060
crgolden
  • 141
  • You can take the integration of $ \frac x{2^x}$ from 1 to $\infty $ – Jasser Sep 20 '14 at 05:54
  • 1
    Dear deeprog use the ratio test. $lim\frac{a_{i+1}}{a_{i}}=\frac{1}{2}$ as n approaches {\infty}$. – M.R. Yegan Sep 20 '14 at 05:56
  • Partial sums are calculated in this question: http://math.stackexchange.com/questions/129302/sum-of-the-series-sum-i-1n-i1-2i However, showing that a series converges is an easier task than finding its sum. – Martin Sleziak Sep 20 '14 at 08:29
  • Here is another question about the same series: http://math.stackexchange.com/questions/441481/why-does-sum-n-0-infty-fracn2n-converge-to-2. And here is a generalization http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn. – Martin Sleziak Sep 20 '14 at 08:42

3 Answers3

7

Another possible way : consider $$\sum\limits_{i=1}^{\infty}{i}{x^i}=x\sum\limits_{i=1}^{\infty}{i}{x^{i-1}}=x \frac{d}{dx}\Big(\sum\limits_{i=1}^{\infty}{x^i}\Big)=x \frac{d}{dx}\Big(\frac{x}{1-x}\Big)=\frac{x}{(1-x)^2}$$ Now, replace $x$ by $\frac{1}{2}$ to get $2$.

The same procedure would apply to $$\sum\limits_{i=1}^{\infty}\dfrac{i}{a^i}=\frac{a}{(a-1)^2}$$

Claude Leibovici
  • 289,558
3

We have $$\lim_{n\to\infty} \left(\frac{n}{2^n}\right)^{1/n}=\lim_{n\to\infty}\frac{n^{1/n}}{2}=\frac{1}{2},$$ so by the Root Test our series converges. The Ratio Test also works nicely.

André Nicolas
  • 514,336
  • Since your answer uses limit of $\sqrt[n]n$, I thought that it might be useful to add links to some posts about this limit: http://math.stackexchange.com/questions/28348/proof-that-lim-n-rightarrow-infty-sqrtnn-1/, http://math.stackexchange.com/questions/115822/how-to-show-that-lim-n-to-infty-n-frac1n-1 – Martin Sleziak Sep 20 '14 at 08:34
1

One way is this:

  1. Prove that for all $i\ge 1$, $i<(1.9)^i$.

  2. Comparison test with $\sum \frac{(1.9)^i}{2^i}=\sum (\frac{1.9}{2})^i$, a convergent geometric series.

vadim123
  • 83,937