How can you solve a tricky sum without knowing the solution and using induction?

[3029298]

The problem

Calculate the following sum:

<br /> \sum_{n=1}^{\infty}\frac{n}{\left(n+1\right)!}<br />

The solution

I know the solution of the partial sum:

<br /> \sum_{n=1}^{K}\frac{n}{\left(n+1\right)!}=\frac{\left(K+1\right)!-1}{\left(K+1\right)!}<br />

If we take the limit of K\rightarrow\infty the sum is equal to 1.

This solution can then be easily proved by induction.

But my question is: how can you obtain a solution without knowing this answer and proving it by induction?

Any help would be really appreciated!

 

[Dick]

Let f(x)=exp(x)/x and consider the derivative of the taylor series of f(x) evaluated at x=1. It's -1+S where S is your series. Now directly evaluate f'(1). What do you conclude S is??

 

[3029298]

The derivative of the Taylor series you mention, looks like this:

<br /> e\left(\left(x-1\right)-\left(x-1\right)^2+\frac{3}{2}\left(x-1\right)^3-\frac{11}{6}\left(x-1\right)^4+...\right)<br />

I do not see anything emerging from this... :shy:

 

[Dick]

The series expanded around x=0. I.e. f(x)=(1/x)*(1+x+x^2/2!+x^3/3!+...). Multiply the 1/x in and differentiate and put x=1.

 

[3029298]

But then the term in the numerator remains 1. In the series it is n, so this does not lead to the correct answer. We in fact want the sum of:

<br /> \frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{n}{(n+1)!}<br />

 

[Dick]

f(x)=1/x+1+x/2!+x^2/3!+x^3/4!+... f'(x)=-1/x^2+1/2!+2x/3!+3x^2/4!+... Put x=1. Do you see your series now?? The n in the numerator comes from the differentiation.

 

[3029298]

wow, this is very cool!

now I'm interested in how you actually thought of this solution!
you could really help me by showing what strategy you use to solve the series with factorials in them, because those are the ones I have the most difficulties with.

 

[D H]

Calculate the following sum:

<br /> \sum_{n=1}^{\infty}\frac{n}{\left(n+1\right)!}<br />

[noparse]By induction,[/noparse] the sum is equal to 1.

But my question is: how can you obtain a solution without knowing this answer and proving it by induction?

Denote the series as S. Since the terms are all positive, the series is absolutely convergent (if it converges). Add e-2[/tex] to S and rearrange terms (valid iff S converges):<br /> <br /> &lt;br /&gt; \begin{aligned}&lt;br /&gt; S + e-2&lt;br /&gt; &amp;amp;= S + \sum_{n=2}^{\infty}\frac 1 {n!} \\&lt;br /&gt; &amp;amp;= S + \sum_{n=1}^{\infty}\frac 1 {(n+1)!} \\&lt;br /&gt; &amp;amp;= \sum_{n=1}^{\infty}\frac{n+1}{(n+1)!} \\&lt;br /&gt; &amp;amp;= \sum_{n=1}^{\infty}\frac{1}{n!} \\&lt;br /&gt; &amp;amp;= e-1&lt;br /&gt; \end{aligned}&lt;br /&gt;<br /> <br /> Thus S+e-2=e-1[/tex], or S=1.

 

[3029298]

beautiful!

thanks a lot!

 

[Dick]

wow, this is very cool!

now I'm interested in how you actually thought of this solution!
you could really help me by showing what strategy you use to solve the series with factorials in them, because those are the ones I have the most difficulties with.

I thought of it by considering the general term in the expansion of exp(x). x^n/n!. I can get an n in the top by differentiating. But then I get n/n!. I want (n-1)/n!. So I divided by x first. That's all. It's not a very general technique, but if you can recognize a series as 'looking like' a power series, it works.

 

[3029298]

thanks for all this help! I learned a lot today

 

[D H]

But my question is: how can you obtain a solution without knowing this answer and proving it by induction?

Finding a general expression for a partial sum by induction and then finding the limit of this partial sum is a perfectly valid technique. Dick and I both used tricks. The partial sum approach of course involves a "trick" as well -- finding an expression for the dang partial sum.

 

[3029298]

yeah exactly, and since I found the partial sum somewhere on the internet, I prefer a healthier approach :p

 

[3029298]

Oh, this is GREAT! By using this trick I could now do the following exercises as well:

<br /> \sum_{n=0}^\infty\frac{1}{(2n)!}<br />

<br /> \sum_{n=0}^\infty\frac{1}{(2n+1)!}<br />

<br /> \sum_{n=1}^\infty\frac{n}{(2n+1)!}<br />

by considering the sinus hyperbolicus and cosinus hyperbolicus! Thanks a lot for this big eye-opener!

 

[Dick]

Oh, this is GREAT! By using this trick I could now do the following exercises as well:

<br /> \sum_{n=0}^\infty\frac{1}{(2n)!}<br />

<br /> \sum_{n=0}^\infty\frac{1}{(2n+1)!}<br />

<br /> \sum_{n=1}^\infty\frac{n}{(2n+1)!}<br />

by considering the sinus hyperbolicus and cosinus hyperbolicus! Thanks a lot for this big eye-opener!

Atta boy! Go nuts!