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Problem:

Determine if $\sum\limits_{n=1}^{\infty}\frac{1+2+\cdots+n}{2^n}$ converges or diverges.

My attempt:

I'm having a hard time with this one.

  • Trying the ratio test, I'm unable to simplify the expression
  • Trying the root test, I get the nth root of a sum, which I don't know how to simplify
  • Trying the integral test, I'm having a hard time performing the actual integration
  • Trying the (limit) comparison test, I can't find any reasonable expression to compare it with, but that might be due to the fact that I've been working on sequences like this for less than a day

Any help appreciated!

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    $\begingroup$ You can expand the numerator into (n(n+1))/2. $\endgroup$ – Tetramputechture Feb 26 '15 at 21:00
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Notice that $1+\dots+n\leqslant n^2$, then you can conclude convergence with the ratio test.

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  • $\begingroup$ Would this be a combination of the comparison test and the ratio test? I can probably conclude convergence for $n^2 / 2^n$, which is an upper bound on the sequence in the problem. $\endgroup$ – Alec Feb 26 '15 at 21:00
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    $\begingroup$ Yes, it is a combination. $\endgroup$ – Davide Giraudo Feb 26 '15 at 21:00
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Hint:You can write the numerator $1+2+\dots+n$ as $\frac{n(n+1)}{2}$

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  • $\begingroup$ I'm literally ashamed! $\endgroup$ – Alec Feb 26 '15 at 21:01
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We can actually evaluate the sum using the binomial theorem (and negative binomial coefficients): $$ \begin{align} \sum_{n=1}^\infty\frac{1+2+3+\cdots+n}{2^n} &=\sum_{n=1}^\infty\frac{\binom{n+1}{2}}{2^n}\\ &=\sum_{n=1}^\infty\frac{\binom{n+1}{n-1}}{2^n}\\ &=\sum_{n=1}^\infty(-1)^{n-1}\frac{\binom{-3}{n-1}}{2^n}\\ &=\frac12\sum_{n=0}^\infty(-1)^n\frac{\binom{-3}{n}}{2^n}\\ &=\frac12\left(1-\frac12\right)^{-3}\\[9pt] &=4 \end{align} $$

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define $f(x)=\frac{4x}{(2-x)^3}$ write a Taylor's series to obtain$$f(x)=\sum_{n=1}^{\infty}\frac{n(n+1)}{2^{n+1}}x^n$$which converges on $|x|<2$, therefore your sum is $f(1)=4$.

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One more way is by using infinite GP, denote the required sum as

$S(x) =\sum_{k=0}^{\infty} \frac{k(k+1) x^k}{2}$, we want $S(1/2)$

$\sum_{k=0}^{\infty} x^k =\frac{1}{1-x}, |x|<1$; d.w.r.t x on both sides to get

$\sum_{k=0}^{\infty} k x^k =\frac{x}{(1-x)^2}=f(x)$

Again d.w.r.x on both sides to get

$\sum_{k=0}^{\infty} k^2 x^k = x f'(x).$

$S(x)=[f(x)+x f'(x)]/2=\frac{x}{(1-x)^3}.$

Finally we get $S(1/2)=4.$

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