Get the solution to "Find the sum of the convergent series. ∑n=1∞19n2+3n−2"

Marvin Mccormick

Answered question

2020-10-25

Find the sum of the convergent series.

\sum_{n = 1}^{\infty} \frac{1}{9 n^{2} + 3 n - 2}

delilnaT

Skilled2020-10-26Added 94 answers

Solution:
Consider the given series is

\sum_{n = 1}^{\infty} \frac{1}{9 n^{2} + 3 n - 2}

Factorise denominator

9 n^{2} + 3 n - 2 = 9 n^{2} + 6 n - 3 n - 2 = (3 n - 1) (3 n + 2)

therefore, the series will be like

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

or,

\sum_{n = 1}^{\infty} \frac{1}{(3 n - 1) (3 n + 2)} = \sum_{n = 1}^{\infty} \frac{1}{9 (n - \frac{1}{3}) (n + \frac{2}{3})}

This is the form of Telescoping series of

a_{n} = n - \frac{1}{3}

and

a_{n + 1} = n + \frac{2}{3}

By telescoping series result

\sum_{n = 1}^{\infty} (\frac{1}{n + k}) (\frac{1}{n + k + 1}) = \frac{1}{k + 1}

Applying these result to given series,

\sum_{n = 1}^{\infty} \frac{1}{9 (n - \frac{1}{3}) (n + \frac{2}{3})} = \frac{1}{9} \sum_{n = 1}^{\infty} \frac{1}{(n - \frac{1}{3}) (n + \frac{2}{3})} = \frac{1}{9} \times \frac{1}{\frac{2}{3}} = \frac{3}{18} = \frac{1}{6}

Hence, the sum of the given convergent series is

\frac{1}{6}

Jeffrey Jordon

Expert2021-12-25Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?