1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.