"How can I evaluate ∑n=1∞2n3n+1" was answered by our community

jamessinatraaa

Answered question

2022-01-19

How can I evaluate

\sum_{n = 1}^{\infty} \frac{2 n}{3^{n + 1}}

Becky Harrison

Beginner2022-01-19Added 40 answers

No need to use Taylor series, this can be derived in a similar way to the formula for geometric series. Lets

Joseph Fair

Beginner2022-01-20Added 34 answers

Consider the generating function

g (x) = \sum_{n = 0}^{\infty} (\begin{matrix} n + k - 1 \\ n \end{matrix}) x^{n} = \frac{1}{{(1 - x)}^{k}}

If we let k=2, then

\sum_{n = 0}^{\infty} (\begin{matrix} n + 1 \\ n \end{matrix}) x^{n} = \frac{1}{{(1 - x)}^{2}}

Since

(\begin{matrix} n + 1 \\ n \end{matrix}) = (n + 1)

we can conclude that

\sum_{n = 0}^{\infty} (n + 1) x^{n} = \frac{1}{{(1 - x)}^{2}}

alenahelenash

Expert2022-01-24Added 556 answers

To avoid differentiating an infinite sum. We start with the standard finite evaluation:

1 + x + x^{2} + . . . + x^{n} = \frac{1 - x^{n + 1}}{1 - x}, | x | < 1

(1) Then by differentiating (1) we have

1 + 2 x + 3 x^{2} + . . . + n x^{n - 1} = \frac{1 - x^{n + 1}}{(1 - x)^{2}} + \frac{- (n + 1) x^{n}}{1 - x}, | x | < 1

(2) and by making

n \to + \infty

in (2), using

| x | < 1

, gives

\sum_{n = 0}^{\infty} (n + 1) x^{n} = \frac{1}{(1 - x)^{2}}

(3)

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