Sum of an infinite series.

Question

I have found a series whose $n$-th term is denoted by $\dfrac{n}{a^n}$. Here $a$ is a constant. I tried to find the sum but failed. Is there any formula for its infinite sum or just an approximation?

Answer 1

The limit can indeed be calculated explicitly. Consider for $|x| < 1$ the function $$f(x) = \sum \limits_{n = 0}^\infty x^n = \frac{1}{1 - x}$$ Differentiating the series termwise and multiplying by $x$ yields $$\frac{x}{(1 - x)^2} = xf'(x) = \sum \limits_{n = 1}^\infty nx^n$$ Now you only need to plug in $x = \frac{1}{a}$. Note that the series obviously diverges for $|x| \ge 1$.

Answer 2

Let $$S(x)=\sum_{i=0}^{N-1} x^n,$$ which equals $$\frac{x^N-1}{x-1}$$ for any $x\ne1$.

Then by deriving term-wise and multiplying by $x$,

$$xS'(x)=\sum_{n=0}^{N-1} nx^n=x\frac{Nx^{N-1}(x-1)-(x^N-1)}{(x-1)^2}=x^N\frac{(N-1)x-N}{(x-1)^2}+\frac x{(x-1)^2}.$$

When $|x|<1$, the first term vanishes for $N\to\infty$.

Answer 3

notice, we have $$T_n=\frac{n}{a^n}$$ $$S_n=\sum_{n=1}^{n} T_n=\sum_{n=1}^{n}\frac{n}{a^n}$$ $$S_n=\frac{1}{a}+\frac{2}{a^2}+\frac{3}{a^3}+\frac{4}{a^4}+\ldots +\frac{n}{a^n}\tag 1$$ Multiplying by $\frac{1}{a}$ & rewriting as follows
$$\frac{1}{a}S_n= \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{a^2}+\frac{2}{a^3}+\frac{3}{a^4}+\frac{4}{a^5}+\ldots +\frac{n-1}{a^{n}}+\frac{n}{a^{n+1}}\tag 2$$ Now, subtracting the corresponding terms of (2) from (1) column wise, we get $$S_n-\frac{1}{a}S_n=\underbrace{\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+\ldots+\frac{1}{a^n}}_{\text{n-terms in G.P.}}-\frac{n}{a^{n+1}}$$ $$\left(\frac{a-1}{a}\right)S_n=\frac{\frac{1}{a}\left(\frac{1}{a^n}-1\right)}{\frac{1}{a}-1}-\frac{n}{a^{n+1}}$$

$$S_n=\left(\frac{a}{a-1}\right)\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$ Hence, taking the limit as $n\to \infty$, the sum of infinite terms $S_{\infty}$ is given as follows $$S_{\infty}=\lim_{n\to \infty}S_n$$ $$=\lim_{n\to \infty}\left(\frac{a}{a-1}\right)\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$ $$=\left(\frac{a}{a-1}\right)\lim_{n\to \infty}\left\{\frac{a^n-1}{a^n(a-1)}-\frac{n}{a^{n+1}}\right\}$$ $$S_{\infty}=\frac{1}{(a-1)^2}\lim_{n\to \infty}\left\{\frac{a^{n+1}-n(a-1)-a}{a^n}\right\}$$ It is obvious that the series converses for $|a|\leq 1$ otherwise diverges