How do I evaluate the power series ∑n=1∞n9n without using the formula for infinite geometric series?

Check with induction that the partial sum is ∑k=1Nk9k=9−9−N(8N+9)64 Evaluate the limit when N goes to infinity and you get the result.

Notice that xddx11−x=∑i=0∞ixi. So differenciating you get xddx11−x=x(1−x)2 Substituting x=19 you get required answer.

Check with induction that the partial sum is ∑k=1Nk9k=9−9−N(8N+9)64 Evaluate the limit when N goes to infinity and you get the result.

Notice that xddx11−x=∑i=0∞ixi. So differenciating you get xddx11−x=x(1−x)2 Substituting x=19 you get required answer.

Key to "How do I evaluate the power series ∑n=1∞n9n..."

kloseyq

Answered question

2022-01-02

How do I evaluate the power series

\sum_{n = 1}^{\infty} \frac{n}{9^{n}}

without using the formula for infinite geometric series?

Answer & Explanation

poleglit3

Beginner2022-01-03Added 32 answers

Check with induction that the partial sum is

\sum_{k = 1}^{N} \frac{k}{9^{k}} = \frac{9 - 9^{- N} (8 N + 9)}{64}

Evaluate the limit when N goes to infinity and you get the result.

Beverly Smith

Beginner2022-01-04Added 42 answers

Notice that

x \frac{d}{d x} \frac{1}{1 - x} = \sum_{i = 0}^{\infty} i x^{i}

. So differenciating you get

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

Substituting

x = \frac{1}{9}

you get required answer.

karton

Expert2022-01-09Added 613 answers

Denote $A = \sum_{n \geq 1} n 9^{- n}, B = \sum_{n \geq 2} n 9^{- n}, C = \sum_{n \geq 1} 9^{- n}, D = \sum_{n \geq 2} 9^{- n}$ . You have the closed system
$A = B + \frac{1}{9}, 9 B - A = C, C = D + \frac{1}{9}, 9 D = C$
which gives the value of A. Of course, as a byproduct of the computation one also gets the value of the geometric series C.

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