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Geometric series

  • Thread starter Benny
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  • #1
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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?

[tex]
\sum\limits_{k = 0}^N {r^k } = \frac{{1 - r^{N + 1} }}{{1 - r}}
[/tex]

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

  • #2
TD
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That is correct :smile:
 
  • #3
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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?
[tex]
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)!}}}
[/tex]

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:
  • #4
quasar987
Science Advisor
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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.
 
  • #5
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Thanks quasar, I see what you mean.
 

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