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Let $\{a_n\}_{n=1}^{\infty}$ be a sequence defined by $$a_n = \frac{n+1}{2^{n+1}}\left(\sum_{k=1}^n \frac{2^k}{k}\right)$$

Show that the sequence converges and find its limit.

Update: After some computation I see that its limit is 1. Maybe we can use the "squeeze" theorem? I proved that $a_n > 1, \forall n \geq 2$ but I can't find the upper bound. I appreciate all help. Thank you

Joshua Mundinger
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primitiveroot
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  • Have you considered using the Monotone Convergence Theorem? – Joshua Mundinger Nov 06 '14 at 15:28
  • I've considered it, but after that I can't find any relation between $a_n$ and $a_{n+1}$ that doesn't involve $n$. Therefore no solution for its limit (that's what I think). – primitiveroot Nov 06 '14 at 15:51
  • Look what I used, do you know this property of diverging series of positive terms? It's often very useful when the general term is of this type. – mvggz Nov 06 '14 at 16:02
  • At class we don't study much about diverging series. That's a nice property! Thanks a lot! – primitiveroot Nov 06 '14 at 16:26
  • it's really easy to prove , you just have to use : $u_n$ ~ $v_n$ <=> $ u_n = v_n +o(v_n) $ You translate this into inequality with an $\epsilon$ : $ u_n - v_n \leq \epsilon*v_n $ for n>N, a certain integer. You sum this from N to n, and you can conclude. It's very useful – mvggz Nov 06 '14 at 16:43
  • Approximate your sum using the exponential integral, then see the section on “asymptotics”. – Lucian Nov 06 '14 at 17:00
  • Intuitively, it seems reasonable to say that $\displaystyle\sum_{k=1}^n \frac{2^k}{k}$ is in the order of $\displaystyle\sum_{k=1}^n \frac{2^k}{n/2}$, but a rigorous proof escapes me. – Lucian Nov 06 '14 at 17:59
  • See here for answer, including Euler-Maclaurin along the lines of @Lucian's suggestion: http://math.stackexchange.com/questions/888354/asymptotic-behavior-of-sum-limits-k-1n-frac2kk/888704#888704 – RRL Nov 06 '14 at 23:06
  • @lucian, that's essentially what I have proved. But it uses few material, like I said to primitiveroot, you only use the definition of $a_n$ ~ $b_n$ and the fact that $\sum a_n$ diverges is the same as : $(\sum a_n) \rightarrow +\infty $. I'll add the proof in my answer – mvggz Nov 07 '14 at 08:41

3 Answers3

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I've got an interesting solution I think, it uses partial sums of diverging series with positive terms:

Let: $b_n = \frac{2^n}{n}$ ; $v_n = b_{n}-b_{n-1}$

$ v_n = \frac{2^n}{n}-\frac{2^{n-1}}{n-1} = \frac{2^{n-1}}{n-1}*[ 1-\frac{2}{n} ] $ ~ $ \frac{2^{n-1}}{n-1} = b_{n-1}$ , when $ n \rightarrow +\infty $

$(b_n) \rightarrow +\infty $, when $n \rightarrow +\infty$ , so $\sum v_n$ diverges and we can say that both partial sums are equivalent:

$ \sum_{k=2}^{n+1} v_k = b_{n+1}-b_1 $ ~ $b_{n+1}$ ~ $ \sum_{k=1}^n b_k $ , when $ n \rightarrow +\infty $

Hence you get: $ a_n = \frac{1}{b_{n+1}}*\sum_{k=1}^n b_k $ ~ 1 , when $ n \rightarrow +\infty $

You get the limit :) , which is 1 like you said

Edit: I'll prove the property I used above. Let's have $ \sum u_n$ and $\sum v_n $ two diverging series of positive terms: $ u_n,v_n \geq 0 $ such as: $u_n$ ~ $v_n$

$u_n$ ~ $v_n$ <=> $ u_n = v_n + o(v_n) $

$ u_n = v_n + o(v_n) $ <=> for any $ \epsilon > 0 , n \geq N => u_n -v_n \leq \frac{\epsilon}{2}*v_n $

Now we sum this from N+1 to n :

$ \sum_{k =N+1}^n u_k -\sum_{k =N+1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =N+1}^n v_k $

=> $ \sum_{k =1}^n u_k -\sum_{k =1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k + \sum_{k =1}^N u_k -(\frac{\epsilon}{2} +1)*\sum_{k =1}^N v_k $

=> $ \sum_{k =1}^n u_k -\sum_{k =1}^n v_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k + \sum_{k =1}^N u_k $ , since $ (\frac{\epsilon}{2} +1)*\sum_{k =1}^N v_k \geq 0 $

Now I'll define : $U_n = \sum_{k =1}^n u_k$, $ V_n = \sum_{k =1}^n v_k $

The hypothesis is : $(V_n) \rightarrow +\infty$ ; so : $ \sum_{k =1}^N u_k =o(V_n) $

=> $ \epsilon > 0 , n \geq N_o => \sum_{k =1}^N u_k \leq \frac{\epsilon}{2}*\sum_{k =1}^n v_k $

You get : $ \epsilon >0 , n > max(N, N_o) => U_n -V_n \leq \epsilon*V_n $

That's exactly : $ U_n -V_n = o(V_n) $ ie $U_n$ ~ $V_n$

mvggz
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By Stoltz-Cesaro

$$\frac{\sum_{k=1}^n \frac{2^k}{k}}{\frac{2^{n+1}} {n+1}}$$

we obtain

$$\frac{\frac{2^{n+1}}{n+1}}{\frac{2^{n+2}}{n+2}-\frac{2^{n+1}}{n+1}}=\frac1{\frac{2(n+1)}{n+2}-1}\to 1$$

Refer to the related

user
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  • This is basically a copy-paste of your previous answer, and would probably be better as a comment. – Arnaud D. Nov 28 '19 at 15:55
  • @ArnaudD. This is not a copy paste, I've indeed write down all the solution and also indicated the general case as related. I'm wondering how is possible that this simple derivation wasn't already indicated as answer. – user Nov 28 '19 at 18:09
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After the change of summation index $j:=n+1-k$, we obtain that $$a_n=\sum_{j=1}^n2^{-j}+\sum_{j=1}^n\frac{j2^{-j}}{n+1-j}.$$ Define $$b_n:=\sum_{j=1}^n\frac{j2^{-j}}{n+1-j}$$ and fix $R\geqslant 1$. Denote $I_R:=\{j\mid n+1-j\geqslant R\}$ and $J_R:=\{j\leqslant n\mid n+1-j\lt R\}$. Since $$\sum_{j\in I_R}\frac{j2^{-j}}{n+1-j}\leqslant \frac 1R\sum_{j=1}^\infty j2^{-j},\mbox{ and }$$ $$\sum_{j\in J_R}\frac{j2^{-j}}{n+1-j}\leqslant\sum_{j\in J_R}j2^{-j}\leqslant n2^{-(n+1-R)}=n2^{-n}2^{R-1},$$ we get $$0\leqslant b_n=\sum_{j\in I_R}\frac{j2^{-j}}{n+1-j}+\sum_{j\in J_R}\frac{j2^{-j}}{n+1-j}\leqslant \frac 1R\sum_{j=1}^\infty j2^{-j}+n2^{-n}2^{R-1}.$$

Davide Giraudo
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