# How can I calculate $\sum_{j=0}^{\infty}(j+1)(\frac{1}{1.05})^{j+1}$

I have to calculate the sum:

$$\sum_{j=0}^{\infty}(j+1)\cdot\left(\frac{1}{1.05}\right)^{j+1}$$

I know it is convergent from the ratio test.

## 1 Answer

Let $$f(x)=\sum_{k=0}^{\infty}(k+1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2-1)(x)^{k+1}=\sum_{k=0}^{\infty}(k+2)x^{k+1}-\sum_{k=0}^{\infty}x^{k+1}$$

for $|x|\lt1$ $$g'(x)=\sum_{k=0}^{\infty}(k+2)x^{k+1}=2x+3x^2+4x^3+\cdots$$ $$g(x)=x^2+x^3+x^4+\cdots=\frac{x^2}{1-x}$$ $$g'(x)=\frac{x(2-x)}{(1-x)^2}$$

$$p(x)=\sum_{k=0}^{\infty}x^{k+1}=x+x^2+x^3+\cdots=\frac{1}{1-x}-1$$

$$f(x)=g'(x)-p(x)$$

$$f(x)=\sum_{k=0}^{\infty}(k+1)(x)^{k+1}=\frac{x(2-x)}{(1-x)^2}-\left(\frac{1}{1-x}-1\right)$$

$$f(x)=\frac{x}{(1-x)^2}$$

Now substitute $x=\dfrac{1}{1.05}$ to get

$$\sum_{k=0}^{\infty}(k+1)\left(\frac{1}{1.05}\right)^{k+1}=420$$