Converging Sum!

Let me show you guys something....

Take $X=\sum _{ i=1 }^{ n }{ \frac { { 2 }^{ i } }{ i } } \Rightarrow { S }_{ n }=\frac { n+1 }{ { 2 }^{ n+1 } } X$

Now, $\Rightarrow { S }_{ n+1 }=\frac { n+2 }{ { 2 }^{ n+2 } } \sum _{ i=1 }^{ n+1 }{ \frac { { 2 }^{ i } }{ i } }$

$\Rightarrow { S }_{ n +1}=\frac { n+2 }{ { 2 }^{ n+2 } } (X+\frac { { 2 }^{ n+1 } }{ n+1 } )=\frac { n+2 }{ { 2 }^{ n+2 } } (\frac { { 2 }^{ n+1 } }{ n+1 } { S }_{ n }+\frac { { 2 }^{ n+1 } }{ n+1 } )$

Put $n\longrightarrow\infty$

$\Rightarrow { S }_{ n+1 }=\frac { { S }_{ n } }{ 2 } +\frac { 1 }{ 2 }$

Since series converges, we must have ${ S }_{ n+1 }={ S }_{ n }$ once $n$ is $\infty$

$\Rightarrow { S }_{ n+1 }={ S }_{ n }=1$

[Can you spot the mistake in this solution? If no, check out the note below.]

Ha! awesome problem,

Let us observe the sequence from the back side, and the truth is far simpler,

$\frac { n+1 }{ { 2 }^{ n+1 } } \left( \frac { 2 }{ 1 } +\frac { 4 }{ 2 } +.....+\frac { { 2 }^{ n-1 } }{ n-1 } +\frac { { 2 }^{ n } }{ n } \right) \quad \\ \\ =\frac { n+1 }{ { 2 }^{ n+1 } } \left( \frac { { 2 }^{ n } }{ n } +\frac { { 2 }^{ n-1 } }{ n-1 } +.....+\frac { 4 }{ 2 } +\frac { 2 }{ 1 } \right) \\ =\frac { 1 }{ 2 } \left( \frac { n+1 }{ n } \right) +\frac { 1 }{ 4 } \left( \frac { n+1 }{ n-1 } \right) .......\\ \\ as\quad n\quad \to\quad \infty \\ \\ =\quad \frac { 1 }{ 2 } +\frac { 1 }{ 4 } ......\quad =\quad \frac { 1 }{ 2 } (\frac { 1 }{ 1-\frac { 1 }{ 2 } } )\quad =\quad 1\quad$

i will use Stolz–Cesàro theorem 1

let $b_{n}=\displaystyle \frac{2^{n+1}}{n+1}$ then $\displaystyle \frac{b_{n+1}}{b_{n}}=\frac{2n+2}{n+2}=2-\frac{4}{n+2}\geq \frac{2}{3}>1 , \forall n\in \mathbb{N}$ or $b_{n+1}>b_{n} , \forall n\in \mathbb{N}$ and $\displaystyle \lim_{n\rightarrow \infty} b_{n}=\lim_{n\rightarrow \infty} \frac{2^{n+1}}{n+1}=\infty$

So , if $\displaystyle a_{n}=\sum_{i=1}^{n} \frac{2^{i}}{i}$ when $\displaystyle \lim_{n\rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}} \\ \\ =\displaystyle \lim_{n\rightarrow \infty} \dfrac{\displaystyle \sum_{i=1}^{n+1} \dfrac{2^{i}}{i} - \sum_{i=1}^{n} \dfrac{2^{i}}{i} }{ \dfrac{2^{n+2}}{n+2}-\dfrac{2^{n+1}}{n+1} } \\ =\displaystyle \lim_{n\rightarrow \infty} \dfrac{\displaystyle \dfrac{2^{n+1}}{n+1} }{ 2^{n+1} \left( \dfrac{n}{(n+1)(n+2)} \right) } \\ =\displaystyle \lim_{n\rightarrow \infty} \dfrac{2^{n+1}}{n+1}\dfrac{(n+1)(n+2)}{2^{n+1}n}$ $=\displaystyle \lim_{n\rightarrow \infty} \dfrac{n+2}{n}=1$

we can use Stolz–Cesàro theorem 1 we get

$\displaystyle \lim_{n\rightarrow \infty} \dfrac{a_{n}}{b_{n}}=\displaystyle \lim_{n\rightarrow \infty} \dfrac{\displaystyle \sum_{i=1}^{n}\dfrac{2^{i}}{i}}{\displaystyle \dfrac{2^{n+1}}{n+1}} \\ = \displaystyle \lim_{n\rightarrow \infty} \dfrac{n+1}{2^{n+1}}\sum_{i=1}^{n}\dfrac{2^{i}}{i} = 1$

We can show (by induction) that : $\displaystyle\sum_{k=0}^n\frac1{\binom{n}{k}}=\frac{n+1}{2^{n+1}}\displaystyle \sum_{k=1}^{n+1}\frac{2^k}{k}$

So $\displaystyle\sum_{k=0}^n\frac1{\binom{n}{k}}=S_n +\frac{n+1}{2^{n+1}} \cdot \frac{2^{n+1}}{n+1}=S_n+1$

We still need to manage this sum. But for $2\le k\le n-2$ , $\binom{n}{k}\ge\binom{n}{2}$ or $\frac{1}{\binom{n}{k}} \leq \frac{1}{\binom{n}{2}} = \frac{2}{n(n-1)}$ one has for $n\ge 4$ ,

$2 \leq \displaystyle \sum_{k=0}^{n} \frac{1}{\binom{n}{k}} \leq 2 + \frac{2}{n} + \frac{2(n-3)}{n(n-1)} \xrightarrow[n\to\infty]{} 2$

We easily get the answer for $S_n$ .

Replace 2 with x>1. $S = \frac{(n+1)\sum_{i=1}^n x^{i}/i}{x^{n+1}}$

Applying Lhopital's rule $\frac{\sum_{i=1}^n x^{i-1}}{x^{n}}$ Which is equal to $\frac{x^{n}-1}{(x-1)x^{n}}$

Which tends to 1/(x-1)

Let $a_i=2^i, b_i=i$ . As n tends to infinity, by stolz cesaro, the limit is $\frac{n+1}{2^{n+1}}* \frac{(2^{n+1})}{n+1}=1$ . ( As sum i=1 to n+1 2^i/I - sum i=1 to n 2^i/i = 2^n+1/n+1)