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Consider the sequence $u_n=\sum_{r=1}^n\frac{r}{2^r},n\geq 1$. Then the limit of $u_n$ as $n\to\infty$ is

$A.$ $1$

$B.$ $2$

$C.$ $e$

$D.$ $\frac12$

I tried solving this mcq question using sandwich theorem, but was unsuccessful, on getting same limits on both sides of the inequality, I built. But then I observed, if $S_n=\sum_{r=1}^n\frac{r}{2^r},$ then, we can claim, $T_n=\sum_{r=1}^n \frac{1}{2^r}\leq S_n.$ As, $S_n$ is convergent, so we have, $\lim_{n\to\infty}T_n=2\leq \lim_{n\to\infty}S_n.$ This is how, I came under the conclusion, $A,D$ are incorrect options. But now, I am confused whether the correct option is $2$ or $e$?

Arthur
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2 Answers2

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So, we can actually calculate this explicitly using a commonly known trick. Consider the sum $1/(1-x) = \sum_{i =0}^{\infty} x^n$, which is "well behaved" in the interval $(-1, 1)$, by which I mean we can commute the derivative with the limit. We have that $x \dfrac{d}{dx}[ 1/(1-x)] = \sum_{i=1}^{\infty} nx^{n}$. I will leave the rest to you, but this is really common trick to finding infinite sums like this, so it is good to remember.

mildboson
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    Please do not answer duplicates. My comment above appeared much before you posted this answer. – Kavi Rama Murthy May 05 '23 at 07:19
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    It didn't appear on my screen when I posted this, otherwise I wouldn't have. In fact, you posted 5 or 6 minutes before me, so "much" is perhaps an exaggeration. Also, aren't we, as posters, trying to not give full solutions just guidance... You linked a complete solution, isn't that considered "inappropriate" for this forum? – mildboson May 05 '23 at 10:48
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In general, for any $|a|<1$, we have $$ \begin{aligned} \sum_{r=1}^n r a^r & =a \sum_{r=1}^n r a^{r-1} \\ & =a \frac{d}{d a}\left(\sum_{r=1}^n a^r\right) \\ & =a \frac{d}{d a}\left(\frac{1}{1-a}\right) \\ & =\frac{a}{(1-a)^2} \end{aligned} $$ In particular, when $a=\frac{1}{2}$, we get $$ \sum_{r=1}^n \frac{r}{2^r}=\frac{\frac{1}{2}}{\left(1-\frac{1}{2}\right)^2}=2 $$

Lai
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