# Geometric series

This has been bothering me for a while. I've seen many different versions of this and I'd just like to get the following cleared up. Is the following true?

$$\sum\limits_{k = 0}^N {r^k } = \frac{{1 - r^{N + 1} }}{{1 - r}}$$

There are other related things I am slightly worried about but I can't remember what they are at the moment so I might post them later. In the mean time, any verification of the above formula that I have quoted would be good thanks.

## Answers and Replies

TD
Homework Helper
That is correct Ok thanks TD. I have another question and I realise that it usually depends on the convention adopted by the particular lecturer but other input shouldn't hurt. When we refer the the nth partial sum of a series. Do we take n >= 1? I mean with a series like $$\sum\limits_{n = 1}^\infty {a_n }$$ it makes sense to take n>=1 but it seems to get awkward when the summation is started at zero instead.

Anyway suppose that when we refer to the nth partial sum of a series we take n>=1 and that the Nth partial sum of the series I stated above is $$s_N = 1 + \frac{1}{{5^N }}$$. Then is the Nth partial sum of the series $$\sum\limits_{n = 2}^\infty {a_n }$$ (I have replaced 1 by 2 as the starting index) $$s_N = \left( {1 + \frac{1}{{5^N }}} \right) + a_{N + 1}$$. I just want to make sure of these things, I always screw up on relatively easy questions due to little technicalities.

Just one more question. If I recall correctly then: $$f\left( x \right) = \sum\limits_{n = 0}^\infty {c_n \left( {x - a} \right)^n } \to f'\left( x \right) = \sum\limits_{n = 1}^\infty {c_n n\left( {x - a} \right)^{n - 1} }$$

So in differentiating the series above, an index shift was required. Ok but what about the following series?
$$f\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n x^{2n + 1} }}{{\left( {2n + 1} \right)!}}} \to f'\left( x \right) = \sum\limits_{n = 0}^\infty {\frac{{\left( { - 1} \right)^n x^{2n} }}{{\left( {2n} \right)!}}}$$

I find that only the first series that I quoted requires an index shift when I differentiate whereas other ones that I've encountered do not. Is there a reason for this?(I am aware of the first term in the first series being a constant and hence disappearing upon differentiation) Is it usually true that if you have a series for a function and you differentiate it inside the interval of convergence then no index shifts are required? Any help would be great.

Last edited:
quasar987
Science Advisor
Homework Helper
Gold Member
An index change is never necessary; always optional. When differentiating power series, it is customary to change the index from 0 to 1 because the oth term is 0 anyway.

Thanks quasar, I see what you mean.