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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

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  • $\begingroup$ Have you considered using the Monotone Convergence Theorem? $\endgroup$ – Joshua Mundinger Nov 6 '14 at 15:28
  • $\begingroup$ 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). $\endgroup$ – primitiveroot Nov 6 '14 at 15:51
  • $\begingroup$ 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. $\endgroup$ – mvggz Nov 6 '14 at 16:02
  • $\begingroup$ At class we don't study much about diverging series. That's a nice property! Thanks a lot! $\endgroup$ – primitiveroot Nov 6 '14 at 16:26
  • $\begingroup$ 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 $\endgroup$ – mvggz Nov 6 '14 at 16:43
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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 $U_n$ ~ $V_n$

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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 $$0\leqslant b_n=\sum_{j\in I_R}\frac{j2^{-j}}{n+1-j}+\sum_{j\in J_R}\frac{j2^{-j}}{n+1-j}\leqslant \frac 1R\sum_{j=1}^\infty j2^{-j}+n2^{-n}2^{R-1}.$$

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