Deducing a series from another series?

[ZeroXz]

Hi,

How can I deduce a series from another series? Like, for e.g.

Summation of n = 0 to infinity (2^(n-1)) / [(2^(n) + 1) (2^(n-1) + 1) = 2/ 3

How to deduce from the above to get

Summation of n = 1 to infinity (2^(n)) / [(2^(n) + 1) (2^(n-1) + 1) = 1

??

What are the necessary steps to deduce??

Any help please? thank you. sorry I can't figure out how to enter mathematical symbols here?

 

[galactus]

If I understand correctly.

You could list out some terms from each and try noticing a pattern between them.

The first few terms of the first sum are

\(\displaystyle S_{1}=\frac{1}{6}+\underbrace{\frac{1}{6}+\frac{2}{15}+\frac{4}{45}+\cdot\cdot\cdot}_{\text{1/2*S2}}\)

The first few terms of the second are

\(\displaystyle S_{2}=\frac{1}{3}+\frac{4}{15}+\frac{8}{45}+\cdot\cdot\cdot\)

It would appear the first series is half the second, except the first also has a 1/6 term.

So, we have \(\displaystyle \frac{1}{6}+\frac{1}{2}S^{2}=S_{1}\)

Since \(\displaystyle S_{1}\) sums to 2/3, enter it in and get \(\displaystyle 2(\frac{2}{3}-\frac{1}{6})=1\)

and 1 is the sum of the second series.

The only difference is the second one is twice the first. And because the first one starts at index n=0, the first term 1/6 is added to the first one.

The second series is the same as the first except the exponent on the second one is n instead of n-1. Also, the index is n=0 instead of n=1. Other than that............



Is that what you meant?.

 

[ZeroXz]

Omg, that's correct. I understand much clearer with this method of yours.