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As a homework assignment, I'm supposed to find the formula for the nth partial sum of the series $$\sum_{n=1}^\infty \left(\frac{5}{4}\right)^{n-1}$$

In my efforts to solve this problem, I've computed the first four terms of the sequence of partial sums to try and find a pattern:

Partial Sum $\quad\quad\quad$ Value

$S_1$ = 1 $\quad\quad\quad\quad\quad\,\,\,\,\,$ 1

$S_2 = 1+\frac{5}{4}$ $\quad\quad\quad\,\,\,\, \frac{9}{4}$

$S_3 = \frac{9}{4}+\frac{25}{16}$ $\quad\quad\,\,\,\, \,\, \frac{61}{16}$

$S_4 = \frac{61}{16}+\frac{125}{64}$ $\quad\quad\ \frac{369}{64}$

I already know that this is a geometric series and that it diverges. However, I can't seem to figure out the formula $S_n$ for the $n^{th}$ partial sum that depends only on $n$, and not on the previous term (i.e. not the recursive formula). The denominator seems to be $4^{n-1}$ , but I can't find the pattern for the numerator. Any help is appreciated!

justanotherstudent
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    Whole series diverged, but for partial sum you can use geometrical progression first $n$ members sum formula. – zkutch Jul 24 '20 at 22:46
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    This is a Geometric Series. – Mark Viola Jul 24 '20 at 22:46
  • I know that the series is geometric and that it diverges, but I am having trouble finding the specific formula for the nth partial sum. – justanotherstudent Jul 24 '20 at 22:48
  • The formula is in the paragraph named Formula of the wiki article cited above. You just need to adapt a little the start and end indexes. – zwim Jul 24 '20 at 22:51
  • @Mark Viola. offtop of course, but as you are here and working, maybe, you find time and for you will be interesting take look at https://math.stackexchange.com/questions/3768032/formula-for-the-multivariate-cumulative-distribution-function-continuous-case/3768234#3768234 as I finished it just now. Thanks. – zkutch Jul 24 '20 at 22:52
  • There are many existing questions on Math SE which answer this question: e.g. https://math.stackexchange.com/questions/2645248/, https://math.stackexchange.com/questions/1566458/, https://math.stackexchange.com/questions/636308 . If none of these answer your question, please edit your question to explain where your confusion lies. – Xander Henderson Jul 24 '20 at 23:01
  • @XanderHenderson yes, thank you for the links! – justanotherstudent Jul 24 '20 at 23:05

2 Answers2

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hint

Let $$S_p=1+a+a^2+a^3+...a^p$$ then

$$aS_p=1+a+a^2+...+a^p-(1-a^{p+1})$$ $$=S_p-(1-a^{p+1})$$

hamam_Abdallah
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Let

$$\sum_{n=1}^N \left(\frac{5}{4}\right)^{n-1}=\sum_{n=0}^{N-1} \left(\frac{5}{4}\right)^{n}$$

and then use that

$$S_N=\sum_{n=0}^{N-1} a^{n} =1+a+\dots+a^{N-1}\implies aS_N=S_N-1+a^N \implies \dots$$

user
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