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infinite series

sbhatnagar

Active member
Jan 27, 2012
95
Fun Problems! Evaluate the following:

1. \( \displaystyle \sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)

2. \( \displaystyle \sum_{n=1}^{\infty}\frac{1+2+2^2+\cdots+2^{n-1}}{n!}\)

3. \( \displaystyle \sum_{n=1}^{\infty}\frac{(1+3^n)\ln^n(3)}{n!}\)

4. \( \displaystyle \sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)

5.\( \displaystyle \sum_{n=0}^{\infty}\dfrac{\displaystyle 2^n x^{x^{2^n}-1}}{\displaystyle 1+x^{2^n}} \)
 
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Markov

Member
Feb 1, 2012
149
Write first series as $\displaystyle\sum\limits_{n=1}^{\infty }{\sum\limits_{j=1}^{n}{\frac{1}{{{2}^{n-1}}(j-1)!}}}=\sum\limits_{j=1}^{\infty }{\sum\limits_{n=j}^{\infty }{\frac{1}{{{2}^{n-1}}(j-1)!}}},$ we can reverse order of summation because of the positivity of the terms of the double series, so now things are straighforward.

You can do similar stuff on second series. Third series is easy by using the expansion of $e^x.$
 
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Sherlock

Member
Jan 28, 2012
59
2. \( \displaystyle \sum_{n=1}^{\infty}\frac{1+2+2^2+\cdots+2^{n-1}}{n!}\)
$\displaystyle \sum_{n \ge 1}~\sum_{0 \le k \le n-1}\frac{2^k}{n!} = \sum_{n \ge 1}\frac{2^n-1}{n!} = \sum_{n \ge 0}\frac{2^n}{n!}-\sum_{n \ge 0}\frac{1}{n!} = e(e-1).$
 

sbhatnagar

Active member
Jan 27, 2012
95
1. \( \displaystyle \sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)
Solution to Problem 1:

Let \( \displaystyle S=\sum_{n=1}^{\infty}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\).

Let \( \displaystyle t_{n}=\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!} \right)}{2^{n-1}}\)

\( \displaystyle \begin{align*} t_{n+1} &= \frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!}+\dfrac{1}{(n)!} \right)}{2^{n}} \\ t_{n+1} &= \frac{1}{2}\frac{\left( 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots +\dfrac{1}{(n-1)!}+\dfrac{1}{(n)!} \right)}{2^{n-1}} \\ t_{n+1} &= \frac{1}{2}t_n +\frac{1}{2^n n!} \\ \sum_{n=0}^{\infty}t_{n+1} &=\frac{1}{2}\sum_{n=0}^{\infty}t_n + \sum_{n=0}^{\infty} \frac{1}{2^n n!} \\ S & =\frac{S}{2}+\sqrt{e} \\ S &= \boxed{2\sqrt{e}}\end{align*}\)

Try solving the last two problems. They are very tricky. :)
 

Markov

Member
Feb 1, 2012
149
My approach for first series is pretty short, the next step is $\displaystyle\sum\limits_{j = 1}^\infty {\frac{1}{{{2^{j - 1}}(j - 1)!}}\sum\limits_{n = 0}^\infty {\frac{1}{{{2^n}}}} } ,$ then combining the value of both series the result equals $2\sqrt e.$
 

Sherlock

Member
Jan 28, 2012
59
Try solving the last two problems. They are very tricky. :)
Don't post solutions yet. I'm working on them.
(Nod)
Are you sure the last one is correct, though?


---------- Post added at 01:45 AM ---------- Previous post was at 12:00 AM ----------

$\displaystyle \begin{aligned} 3. ~f(x) & = \sum_{n \ge 1}\frac{(1+3^n)x^n}{n!} = \sum_{n \ge 1}\frac{x^n}{n!}+\sum_{n \ge 1}\frac{3^nx^n}{n!} \\& = e^{3x}+e^x-2 \Rightarrow f(\ln{3}) = 28.\end{aligned}$

$\displaystyle \begin{aligned} 5. ~ S & = \sum_{n \ge 0}\frac{2^nx^{2^n-1}}{1+x^{2^n}} = \sum_{n \ge 0}\frac{(1+x^{2^n})'}{1+x^{2^n}} = \bigg[\sum_{n \ge 0}\ln\left(1+x^{2^n}\right)\bigg]' \\& = \bigg[\ln\bigg(\prod_{n \ge 0} (1+x^{2^n})\bigg)\bigg]' = \bigg[\ln\bigg(\frac{1}{1-x}\bigg)\bigg]' =\boxed{\dfrac{1}{1-x}}. \end{aligned}$
 
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sbhatnagar

Active member
Jan 27, 2012
95

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
I wasn't aware that $\displaystyle \prod_{n=0}^{\infty} \left(1+x^{2^{n}} \right) = \sum_{n=0}^{\infty} x^{n} $. But from just writing out terms it appears to be true.
 

Sherlock

Member
Jan 28, 2012
59
I wasn't aware that $\displaystyle \prod_{n=0}^{\infty} \left(1+x^{2^{n}} \right) = \sum_{n=0}^{\infty} x^{n} $. But from just writing out terms it appears to be true.
We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $

But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $

Thus $\displaystyle \frac{\left(1-x\right)}{\left(1-x^{2^{n}}\right)}\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)}{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)} = 1 \implies \prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{1-x^{2^{n+1}}}{1-x}. $

The difference of two squares can do wonders! We get our case when $n \to \infty$ of course.
 

sbhatnagar

Active member
Jan 27, 2012
95

4. \( \displaystyle \sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)
Let \( \displaystyle S=\sum_{n=1}^{n}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} \)

\( \displaystyle S=\sum_{n=1}^{\infty}\dfrac{\displaystyle x^{2^{n-1}}}{\displaystyle 1-x^{2^n}} =\sum_{n=1}^{\infty}\frac{x^{2^{n-1}}+1-1}{(1-x^{2^{n-1}})(1+x^{2^{n-1}})} = \sum_{n=1}^{\infty} \frac{1}{(1-x^{2^{n-1}})}-\frac{1}{(1-x^{2^{n-1}})(1+x^{2^{n-1}})} = \sum_{n=1}^{\infty}\frac{1}{(1-x^{2^{n-1}})}-\frac{1}{(1-x^{2^{n}})}\)

\( \displaystyle S=\frac{1}{1-x}-\frac{1}{1-x^2}+\frac{1}{1-x^2}-\frac{1}{1-x^4}+\cdots= \boxed{\dfrac{1}{1-x}}\)

(Music)
 

melese

Member
Feb 24, 2012
27
We have $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right)}{\left(1-x^{2^{n+1}}\right)}\prod_{1 \le k \le n+1}\left(1-x^{2^{k}}\right) = \frac{\left(1-x\right) }{\left(1-x^{2^{n+1}}\right)}\prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right). $

But $\displaystyle \left(1-x^{2^{k+1}}\right) = \left(1-x^{2^{k}}\right)\left(1+x^{2^{k}}\right)$, thus $\displaystyle \prod_{0 \le k \le n}\left(1-x^{2^{k+1}}\right) =\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right). $

Thus $\displaystyle \frac{\left(1-x\right)}{\left(1-x^{2^{n}}\right)}\prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)}{\prod_{0 \le k \le n}\left(1-x^{2^{k}}\right)} = 1 \implies \prod_{0 \le k \le n}\left(1+x^{2^{k}}\right) = \frac{1-x^{2^{n+1}}}{1-x}. $

The difference of two squares can do wonders! We get our case when $n \to \infty$ of course.
I think there's a simpler reason. When multiplying out, we find that $\displaystyle\prod_{0\leq k\leq n}(1+x^{2^k})$ is the sum of terms of the form $x^m$ (except for $1$), where $m$ is a sum of powers of two. Then we may notice that any $1\leq m\leq2^{n+1}-1$ is obtained exactly once because any number can be written uniquely in base-2 representation.

For example, take $n=3$: $(1+x^{2^0})(1+x^{2^1})(1+x^{2^2})(1+x^{2^3})=1+x^{2^0}+x^{2^1}+x^{2^0+2^1}+x^{2^2}+x^{2^0+2^2}+x^{2^1+2^2}+x^{2^0+2^2+2^2}=1+x+x^2+x^3+x^4+x^5+x^6+x^7$

Therefore, $\displaystyle\prod_{0\leq k\leq n}(1+x^{2^k})=1+\sum_{1\leq m\leq 2^{n+1}-1}x^m$

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