Get the answer to "Sum of the series:∑n=0∞1+n3n"

cooopt392a0i

Answered question

2022-02-24

Sum of the series:

\sum_{n = 0}^{\infty} \frac{1 + n}{3^{n}}

Answer & Explanation

paralovut91

Beginner2022-02-25Added 7 answers

Using the usual geometric series, you have

\frac{x}{1 - x} = \sum_{n = 1}^{\infty} x^{n}

thus differentiating you get

\sum_{n = 1}^{\infty} n x^{n - 1} = \frac{1}{{(1 - x)}^{2}}

thus

\sum_{n = 1}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}}

Then

\sum_{n = 1}^{\infty} n x^{n} = \frac{x}{{(1 - x)}^{2}}

So your sum is just

\sum_{n = 1}^{\infty} \frac{n}{3^{n}} = \frac{\frac{1}{3}}{{(1 - \frac{1}{3})}^{2}} = \frac{3}{4}

Nathan Kent

Beginner2022-02-26Added 8 answers

This is just a different take on (much) earlier answers, mostly to emphasize what can sometimes be gained by playing with the index of summation.
Shifting the index of summation and noting that including

n x^{n - 1}

for

n = 0

adds nothing to the sum, one obtains (for

| x | < 1

)

\sum_{n = 0}^{\infty} (n + 1) x^{n} = \sum_{n = 1}^{\infty} n x^{n - 1}

= \sum_{n = 0}^{\infty} n x^{n - 1} = {(\sum_{n = 0}^{\infty} x^{n})}^{'}

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

Setting

x = \frac{1}{3}

gives

\sum_{n = 0}^{\infty} \frac{n + 1}{3^{n}} = \frac{9}{4}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Calculus 2

Didn't find what you were looking for?