0
$\begingroup$

I need help with this problem:

Find the following infinite sums. (Most of the cases are equal to $f(a)$ where $a$ is an obvious number and $f(x)$ is defined by a power series. To calculate the series, it is necessary to effect the necessary arrangements until there appear some well-known power series.)

  1. $\sum_{n=0}^\infty \ \frac{1}{(2n)!}$
  2. $\sum_{n=0}^\infty \ \frac{n}{2^n}$

For the first one, I really don't know how to begin. For the second one, I started by rewriting it like: $\sum_{n=0}^\infty \ n(\frac{1}{2})^n$ so it looks like a geometric series; after that I don't know what to do. Is it ok if I divide it by n and end up with $\sum_{n=0}^\infty \ (\frac{1}{2})^n=\frac{1}{1-\frac{1}{2}}=2$?

$\endgroup$
2
$\begingroup$

Note that $\cosh(z)=\sum_{n=0}^{\infty}{\frac{1}{(2n)!}z^{2n}}$. So, $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\cosh(1)$$

For the second, if $|z|<1$, then $\sum_{n=0}^{\infty}{z^n}=\dfrac{1}{1-z}$, I suppose you know that. Therefore, $$\sum_{n=1}^{\infty}{nz^{n-1}}=\dfrac{1}{(1-z)^2}\implies \sum_{n=0}^{\infty}{nz^{n}}=\dfrac{z}{(1-z)^2}$$ Now, take $z=\frac{1}{2}.$

I will explain the first sum using exponential. We have $\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}$, each time you see factorial is a good idea think in exponential series. So, note that in this series you have only "even factorials", So I'll try to form this series. Note that $$e^{x}+e^{-x}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}}+\sum_{n=0}^{\infty}{\frac{1}{n!}(-1)^{n}x^{n}}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}(1+(-1)^n)}$$

Clearly, the terms $1+(-1)^n=\{0,2\}$ if $n$ is even or odd number. so, we only considered the even numbers, i.e., $$e^{x}+e^{-x}=\sum_{n=0}^{\infty}{\frac{1}{n!}x^{n}(1+(-1)^n)}=\sum_{n=0}^{\infty}{\frac{2}{(2n)!}x^{n}}$$

You can see that this looks a lot like the series we are looking for. Now take $x=1$ in the above,

$$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\frac{1}{2}(e^{1}+e^{-1})$$

And using the complex form, $e^{ix}=\cos(x)+i\sin(x)$ where $i=\sqrt{-1}$, you can see $$e^{1}+e^{-1}=e^{i(-i)}+e^{i(i)}=(\cos(-i)+i\sin(-i))+(\cos(i)+i\sin(i))$$ $$=(\cos(i)-i\sin(i))+(\cos(i)+i\sin(1))=2\cos(i)$$ And we "define", $\cos(i)=\cosh(1)$. Finally, $$\sum_{n=0}^{\infty}{\frac{1}{(2n)!}}=\frac{1}{2}(e^{1}+e^{-1})=\cos(i)=\cosh(1)$$

$\endgroup$
  • $\begingroup$ How do you know that? I had no idea that $\cosh(x)$ was equal to that. I think that I'm not suppose to now that, because that series for $\cosh(x)$ isn't anywhere in my book. $\endgroup$ – davidllerenav Feb 11 at 4:32
  • $\begingroup$ I saw the answer on the book, and it used $\e^x$ and $\e^-x$, but I don't understand it. $\endgroup$ – davidllerenav Feb 11 at 4:33
  • $\begingroup$ @davidllerenav use the exponential form of the $\cosh$ function. $\endgroup$ – aleden Feb 11 at 4:42
  • 1
    $\begingroup$ cosh($x$)=$(e^x+e^{-x})/2$ $\endgroup$ – J. W. Tanner Feb 11 at 4:49
  • 1
    $\begingroup$ In general we can write $$\sum_{n=0}^N a_{2n}=\frac12\sum_{n=0}^{2N} (1+(-1)^n)a_n$$Setting $a_n=\frac1{n!}$ and letting $N\to \infty$, we find that $$\sum_{n=0}^\infty \frac1{(2n)!}=\frac12\left(\sum_{n=0}^\infty \frac1{n!}+\sum_{n=0}^\infty \frac{(-1)^n}{n!}\right)=\frac12(e+e^{-1})$$ $\endgroup$ – Mark Viola Feb 11 at 6:01
1
$\begingroup$

For the second sum, $$f(x)=\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$ $$f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$$ $$\frac{x}{(1-x)^2}=\sum_{n=0}^\infty nx^n$$ For $x=\frac{1}{2}$, $$\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}=2=\sum_{n=0}^\infty \frac{n}{2^n}$$

$\endgroup$
  • $\begingroup$ You multiplied $f'(x)=\sum_{n=0}^\infty nx^{n-1}=\frac{1}{(1-x)^2}$ by x, right? That's why we end up with $\frac{x}{(1-x)^2}=\sum_{n=0}^\infty nx^n$? $\endgroup$ – davidllerenav Feb 11 at 4:50
  • $\begingroup$ @davidllerenav Yep, the crucial step is taking the derivative in order to get the $n$ term as a coefficient in the power series. $\endgroup$ – aleden Feb 11 at 4:59
  • $\begingroup$ Ok, thanks. I understand that. $\endgroup$ – davidllerenav Feb 11 at 5:36
1
$\begingroup$

For the first problem : $$\cos x = 1- \frac{x^2}{2!} + \frac{x^4}{4!}-\frac{x^6}{6!}......$$ Put the value of $x$ as $i$ to get: $$\cos (ix) = 1- \frac{(ix)^2}{2!} + \frac{(ix)^4}{4!}-\frac{(ix)^6}{6!}......$$ Solve further to get: $$\cos (ix) = 1+ \frac{x^2}{2!} + \frac{x^4}{4!}+\frac{x^6}{6!}......$$ Put the value of $x=1$ to get $$\cos i = 1+ \frac{1}{2!} + \frac{1}{4!}+\frac{1}{6!}......=\sum_{n=0}^\infty \ \frac{1}{(2n)!}$$ The second series is an arithmetico geometric series with the $n^{th}$ term as: $$T_n=\frac{n}{2^n}$$ when you calculate the sum(start the sum from n=1 as it wont matter): $$S_{\infty}= \frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}.....$$ You can see that the numerators are in AP where as the fractions are in GP. Just multiply the above expression with the common ratio i.e. $\frac{1}{2}$ to get $$\frac{1}{2}S_{\infty}=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\frac{4}{2^5}...$$ Subtract both of these equations to get : $$\frac{1}{2}S_{\infty} = \frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}...$$ Use the formula for infinite GP in the RHS to get : $$\frac{1}{2}S_{\infty} = \frac{\frac{1}{2}}{1-\frac{1}{2}}$$ You will get $S_{\infty}=2$.

PS: In the first summation you can also use $\cos hx$ instead of $\cos ix$ because: $$\cos hx = \cos ix = \frac{e^{i.(ix)}+e^{-i.(ix)}}{2}=\frac{e^{x}+e^{-x}}{2}$$ Hope this helps .....

$\endgroup$
1
$\begingroup$

Observe that for any $x \in \Bbb R$ the Taylor series expansion of $e^x$ about $x=0$ is $$\sum_{n=0}^{\infty} \frac {x^n} {n!}.$$ Note that $$\begin{align} e+e^{-1} & = 2 \sum_{n=0}^{\infty} \frac {1} {(2n)!} \\ \implies \sum_{n=0}^{\infty} \frac {1} {(2n)!} & = \frac {e+e^{-1}} {2} =\cosh (1). \end{align}$$

For the second one observe that $$\begin{align} \sum_{n=0}^{\infty} \frac {n} {2^n} & = \sum_{n=1}^{\infty} \frac {n} {2^n} \\ & = \sum_{n=1}^{\infty} \frac {1} {2^n} + \sum_{n=2}^{\infty} \frac {1} {2^n} + \sum_{n=3}^{\infty} \frac {1} {2^n} + \cdots \\ & = 1 + \frac 1 2 + \frac {1} {2^2} + \cdots \\ & = \sum_{n=0}^{\infty} \frac {1} {2^n} \\ & = \frac {1} {1 - \frac 1 2} \\ & = 2. \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.