Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Geometric series

  1. Oct 27, 2005 #1
    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.
  2. jcsd
  3. Oct 27, 2005 #2


    User Avatar
    Homework Helper

    That is correct :smile:
  4. Oct 27, 2005 #3
    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 [tex]\sum\limits_{n = 1}^\infty {a_n } [/tex] 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 [tex]s_N = 1 + \frac{1}{{5^N }}[/tex]. Then is the Nth partial sum of the series [tex]\sum\limits_{n = 2}^\infty {a_n } [/tex] (I have replaced 1 by 2 as the starting index) [tex]s_N = \left( {1 + \frac{1}{{5^N }}} \right) + a_{N + 1} [/tex]. 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: [tex]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} } [/tex]

    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: Oct 27, 2005
  5. Oct 27, 2005 #4


    User Avatar
    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.
  6. Oct 28, 2005 #5
    Thanks quasar, I see what you mean.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook