Sum of n terms and infinite terms where term is $\frac{n}{a^n}$

Question

$\frac{n}{a^n}=T_n$
So $\frac1a+\frac2{a^2}+...\infty=S_\infty$
Multiply $S_\infty$ by $\frac1a$.
$\frac1aS_\infty=\frac1{a^2}+\frac2{a^3}+...\infty$
$S_\infty-\frac1aS_\infty=\frac1a+\frac1{a^2}+...\infty$
$S_\infty=\frac{\frac1a}{1-\frac1a}\frac{a}{a-1}$
$S_\infty=\frac a{(a-1)^2}$
Is it right? And I cant quite figure out for the limited sum (till $n$). Please show how.
Edit: $x$ doesn't exist in the interval $[-1,1]$

The final result is correct, but you should be a bit careful when calculating series. Do you have any condition on $a$ (you should)? — mickep, Aug 12 '15 at 11:19
A few related questions: http://math.stackexchange.com/questions/87030/proof-by-induction-sum-limits-i-0n-i-2i-1-n1-2n-1 http://math.stackexchange.com/questions/11464/how-to-compute-the-formula-sum-r-1d-r-cdot-2r http://math.stackexchange.com/questions/90637/what-is-the-limit-of-sum-limits-n-1-inftyn2-3n http://math.stackexchange.com/questions/30732/how-can-i-evaluate-sum-n-0-infty-n1xn — Martin Sleziak, Aug 12 '15 at 12:40

score 2 · Accepted Answer · answered Aug 12 '15 at 11:22

2

Your derivation is correct if you have proven that the series is in the first place convergent. If not, then $S_\infty$ does not even exist and you cannot reason about its value. For now, let us ignore this issue.

To get the sum of the first $n$ terms, just do exactly the same technique.

$S_n = \frac{1}{a} + \frac{2}{a^2} + \cdots + \frac{n}{a^n}$.

$\frac{1}{a} S_n = \frac{1}{a^2} + \frac{2}{a^3} + \cdots + \frac{n}{a^{n+1}}$.

$S_n - \frac{1}{a} S_n = ( \frac{1}{a} + \frac{1}{a^2} + \cdots + \frac{1}{a^n} ) - \frac{n}{a^{n+1}}$.

[You should know how to simplify the right-hand side and hence get the answer.]

Note that this technique does not face the same convergence issues because it is a finite sum. You can use the result you get for the finite sum and see what is the limit as $n \to \infty$, in order to avoid convergence issues in your original method for the infinite sum.

answered Aug 12 '15 at 11:22

user21820

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Geometric sequence? – Aditya Agarwal Aug 12 '15 at 11:24
@AdityaAgarwal: Yes but notice you can derive the formula for the sum of a finite geometric sequence yourself using exactly the same technique! – user21820 Aug 12 '15 at 11:24
I know! Good observation though (y). Aint it be a complicated one though (For my question) – Aditya Agarwal Aug 12 '15 at 11:27
@AdityaAgarwal: It is not at all complicated. Note that this method is the most general possible and the formula for the finite sum works not just for $a \in \mathbb{R} \smallsetminus {0}$ but for any field $F$ and $a \in F \smallsetminus {0}$. – user21820 Aug 12 '15 at 11:31
Can you please show the formula? – Aditya Agarwal Aug 12 '15 at 11:40
@AdityaAgarwal: Why don't you try first and tell me the formula you get, and I'll check for you? – user21820 Aug 12 '15 at 11:51
Ohk. But Ain't it a arithmetico-geometric series? – Aditya Agarwal Aug 13 '15 at 08:30
@AdityaAgarwal: What are you talking about? You should know how to simplify the right-hand side of the last equation in my answer, which will give you the answer. – user21820 Aug 13 '15 at 09:57
https://en.wikipedia.org/wiki/Arithmetico-geometric_sequence – Aditya Agarwal Aug 13 '15 at 09:58
@AdityaAgarwal: Well yes. Anyway the method I've shown you already gets you the formula, and similarly if you have a higher degree polynomial multiplied by a geometric progression, so there is no need to memorize or whatever. – user21820 Aug 13 '15 at 10:00
$S_n=\frac{a^{n+1}-a-na+n}{a^n(a-1)^2}$? – Aditya Agarwal Aug 13 '15 at 10:21
@AdityaAgarwal: Correct. Note that the method requires $a \ne 1$. What happens in the edge case of $a = 1$? – user21820 Aug 13 '15 at 10:40
Infinitttttte!!! – Aditya Agarwal Aug 13 '15 at 10:42
@AdityaAgarwal: No. It just means you get $0 S_n = 0$ so you cannot get $S_n$ this way, and have to use a different method. In this case you want $1+2+\cdots+n$, which I'm sure you know how to get. – user21820 Aug 13 '15 at 10:46
Ohkk, sorry, I was talking for $1+2+3+...+\infty$. – Aditya Agarwal Aug 13 '15 at 11:07
And for that, it would obviously be $\frac{n(n+1)}2$ – Aditya Agarwal Aug 13 '15 at 11:07
@AdityaAgarwal: Ah yup right. It is a common phenomenon that this shifting technique does not work for edge cases, so one should be careful to check that no step is invalid. – user21820 Aug 13 '15 at 11:09

score 1 · Answer 2 · answered Aug 12 '15 at 11:25

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For finite $N$, consider $$\sum_{n=1}^Nne^{sn} = \frac{\partial}{\partial s} \sum_{n=1}^N e^{sn} =\frac{\partial}{\partial s}e^s\frac{1-e^{sN}}{1-e^s} $$

by putting $s=-\ln a$, you will recover your series.

answered Aug 12 '15 at 11:25

mastrok

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score 0 · Answer 3 · answered Aug 12 '15 at 11:25

If I am understanding you correctly, you are wishing to find the closed form of the sum $$ \sum_{n=1}^\infty nx^n$$ Your derivation is incorrect for all $x \geq 1$. To see this in your notation note that $$ S_N = \sum_{n=1}^N nx^n \geq \sum_{n=1}^N n \to \infty, \ N \to \infty$$ Then the "computation" you have done is invalid, because you are multiplying by a number that does not exist! However with $|x| < 1$ your derivation is correct. If you are interested in a more rigorous analysis I suggest differentiating the geometric series $$ \sum_{n=1}^\infty x^n$$ and nothing that for each term $$ \frac{d}{dx} x^n = nx^{n-1}$$ I am sure you can google what I just suggested, as this question has been asked extensively on the internet and on mathstack in particular!

Your so-called more rigorous analysis is in fact not rigorous, since you didn't mention that you made use of a high-power theorem that allows termwise differentiation of a power series. There is completely no need to invoke any real analysis at all, which is not only redundant but also not general. See my answer which shows how to derive the formula for the finite sum that works in any field and not just for $\mathbb{R}$. — user21820, Aug 12 '15 at 11:29

Sum of n terms and infinite terms where term is $\frac{n}{a^n}$

3 Answers3