Question

What is the answer in sum to infinity if n/5^n??

What is the answer in sum to infinity if n/5^n

Answer 1

`sum_(n=k)^oo n/(5^n)=1/(4*5^(k-1))[k+1/4]`

Explanation:

Assuming you meant:

`sum_(n=k)^oo n/(5^n)`

Since you didn't state the lower bound of `n`, I'm going to solve a general case.

If we write our sum out, we have:

`S = k/5^k+(k+1)/(5^(k+1))+(k+2)/(5^(k+2))+...`

This is an arithmetico-geometric sequence. To solve this, we can use a similar method to solving geometric series:

Multiply both sides by `5`, the ratio of the geometric progression in the denominators.

`1/5S=k/color(red)(5^(k+1))+(k+1)/color(red)(5^(k+2))+...`

Notice how the first term has the denominator of the second term in the initial series, the second term has the third one's denominator in the initial series and so on and so forth.

Since we have a lot of common denominators, we can substract the second series from the first to get the following:

`(1-1/5)S =k/(5^k) +(k+1)/(5^(k+1)) - color(red)(k/(5^(k+1)))+(k+2)/(5^(k+2))-color(red)(k+1)/(5^(k+2))+...`

`(1-1/5)S = k/(5^k) + color(blue)(1/(5^(k+1)) + 1/(5^(k+2))+...`

The sum highlighted in blue is a geometric series, for which we have a formula:

`a+at+at^2+at^3+...= a/(1-t)=("first term")/("1-common ratio")`

In our case, `a = 1/(5^(k+1))` and `t=1/5`.

`=> color(blue)(1/(5^(k+1)) + 1/(5^(k+2))+...=(1/(5^(k+1)))/(1-1/5)`

`=5/(4*5^(k+1))=1/(4*5^k)`

Thus, we must have

`4/5S =k(1/(5^k)) + 1/4(1/5^k) = 1/(5^k)[k+1/4]`

`:. S = 5/(4*5^k)[k+1/4]=1/(4*5^(k-1))[k+1/4]`

If, for example, the initial value is `k=1`, then

`S_1 = 1/(4*5^(color(red)(1)-1))[1+1/4]=1/4*5/4=5/16`