Limit of a sum as n approaches infinity.

Question

I have a sum

$$T_n(x)= 1 + 2x + 3x^2 + \dots +n x^{n−1}.$$

and I am supposed to find the limit of $T_n(1/44)$ as $n$ approaches infinity.

I would appreciate any suggestions on how to proceed.

Answer 1

Note that the expression you evaluate is the derivative of $1+ x + x^2 + \dots$. Find a closed form expression for this, and continue from there.

Answer 2

Hint:

$$(1-x)T_n(x) = 1+x+x^2+x^3+\cdots+x^{n-1}-x^n$$

For $x \in (0, 1)$, this gives $$\lim_{n \to \infty} T_n(x) = \frac{1}{(1-x)^2}$$

Answer 3

HINT: Note that $T_n(x)=(F_n(x))'=(1+x+.......+x^n)'$ $n\to\infty$ $F_n(x) \to \frac{1}{1-x} $ if $x<1$.

Answer 4

The question is to find the sum of the series $\sum\limits_{n=1}^{\infty}na^{n-1}$, where $a=1/44$. We know that $$\sum\limits_{n=1}^{\infty}a^n=\frac{a}{1-a}.$$ From Abel formula for summation with $A_n=\sum\limits_{k=1}^n1=n$ we have $$\sum\limits_{n=1}^{\infty}a^n=\sum\limits_{n=1}^{\infty}A_n(a^n-a^{n+1})+\lim\limits_{n\to \infty}A_na^n$$ $$\sum\limits_{n=1}^{\infty}a^n=(1-a)\sum\limits_{n=1}^{\infty}na^{n}=\frac{a}{1-a}\,\,\, (*)$$

From (*) it follows that $$\lim\limits_{n\to \infty}T_n(1/44)=\sum\limits_{n=1}^{\infty}na^{n-1}=\frac{1}{(1-a)^2}=\frac{1936}{1849}.$$