Expert Answer to "How can you derive that∑n=1∞n3n=34 ?"

Ivor Schofield

Answered question

2022-02-23

How can you derive that

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

Aiden Paterson

Beginner2022-02-24Added 1 answers

Let

S_{n} = \sum_{n = 1}^{\infty} \frac{n}{3^{n}}

(1)

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

(1)-(2)

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

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

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

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

Thus,

S_{n} = \frac{3}{4}

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