## Series sum problem

Forum for the GRE subject test in mathematics.
speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Series sum problem

Hi, anyone who knows how to evaluate this sum?

$$\sum_{n=1}^\infty \frac{n}{4^{n+1}}$$

I'm think I'm quite poor in evaluating sums of series of finite sums, since I got stuck most of the time I encounter them, can anyone recommend some material about this topic (it would be great if it focus on methods of evaluating the sum) so that I could study more about this? Thanks

mathQ
Posts: 41
Joined: Thu Mar 25, 2010 12:14 am

### Re: Series sum problem

speedychaos4 wrote:Hi, anyone who knows how to evaluate this sum?

$$\sum_{n=1}^\infty \frac{n}{4^{n+1}}$$

I'm think I'm quite poor in evaluating sums of series of finite sums, since I got stuck most of the time I encounter them, can anyone recommend some material about this topic (it would be great if it focus on methods of evaluating the sum) so that I could study more about this? Thanks

Best would be read about Power Series and Taylor Series. For gre subject test, Princeton book also covers the part of Sequences and Series in a neat way.

Is the ans to above problem 1/9 ?

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Series sum problem

Thanks MathQ, I will grab some materials about those series.
btw, is Fourier series an important topic in the math test?

the ans is correct, but would you mind showing me the process of evaluating this? Thanks.

labyrinth9
Posts: 3
Joined: Tue Apr 06, 2010 4:54 am

### Re: Series sum problem

f(4) = derivative(integral(f(x)))|x=4, since they are inverses.

So, we can represent n/(4^(n+1)) as

d/dx[int[n/x^(n+1)]]|x=4

= d/dx[-(1/x)^n]|x=4

Therefore, we have the ans is equal to

d/dx[SUM[-(1/x)/(1-(1/x))]]|x=4, since d/dx is linear

= 1/(x-1)^2|x=4

= 1/9

ps. Sorry I don't know LaTeX... I really should learn.

speedychaos4
Posts: 47
Joined: Mon Mar 22, 2010 2:42 am

### Re: Series sum problem

labyrinth9 wrote:f(4) = derivative(integral(f(x)))|x=4, since they are inverses.

So, we can represent n/(4^(n+1)) as

d/dx[int[n/x^(n+1)]]|x=4

= d/dx[-(1/x)^n]|x=4

Therefore, we have the ans is equal to

d/dx[SUM[-(1/x)/(1-(1/x))]]|x=4, since d/dx is linear

= 1/(x-1)^2|x=4

= 1/9

ps. Sorry I don't know LaTeX... I really should learn.

Thanks for the nice explanation, I should be able to handle this type of questions now. btw, $$\LaTeX$$ is quite interesting and the basics are to learn! You should read the lshort file if you have time, it is quite helpful.