Calculating the Sum of a series

[Saladsamurai]

\sum_{k=1}^{\infty}(\frac{1}{5^k}-\frac{1}{k(k+1)})

Now by partial fractions and distributing the sum across all three terms I get

\sum_{k=1}^{\infty}\frac{1}{5^k}-\sum_{k=1}^{\infty}\frac{1}{k}+\sum_{k=1}^{\infty}\frac{1}{k+1}

Then I am going with: 1st is geometric; 2nd is Harmonic; and 3rd is similar to Harmonic So, 2nd and 3rd diverge.

So the sum should equal \frac{a}{1-r}.

But this the not match the text answer, what am I doing wrong?

Thanks,
Casey

EDIT: I noticed an example in my text in which I use (1/k)-1/(k+1) to write the closed form and get a finite answer of 1.

Why is this the case if I can distribute the sigma across the terms.

 

[Avodyne]

Your last two series do not completely cancel; write out the first few terms of each to see.

 

[Saladsamurai]

Your last two series do not completely cancel; write out the first few terms of each to see.

I can see now that -\frac{1}{k(k+1)}) goes to 1, but why does distributing change things? Or am I doing something wrong here?

Casey

 

[dynamicsolo]

I can see now that -\frac{1}{k(k+1)}) goes to 1, but why does distributing change things? Or am I doing something wrong here?

Casey

It's not a question of how things are distributed: it's that the two harmonic series are just one term out of step with each other. So the individual series are divergent, but their difference is finite (a classic case of an indeterminate difference yielding a finite non-zero value).

The difference

-\sum_{k=1}^{\infty}\frac{1}{k}+\sum_{k=1}^{\infty} \frac{1}{k+1}

is really like writing

-\sum_{k=1}^{\infty}\frac{1}{k}+\sum_{k=2}^{\infty} \frac{1}{k} ,

so all the terms except the first one in the first sum cancel exactly...

[EDIT: Sorry, moved that index the wrong way before...]

Ah, also beware: the value for a in the geometric series is not 1 (maybe you already caught that, but a frequent mistake made in using the formula for an infinite geometric series is not observing the index for the first term).