# Find the sum of the convergent series. sum_{n=1}^inftyfrac{1}{9n^2+3n-2}

Find the sum of the convergent series.
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{9{n}^{2}+3n-2}$
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Solution:
Consider the given series is $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{9{n}^{2}+3n-2}$
Factorise denominator $9{n}^{2}+3n-2=9{n}^{2}+6n-3n-2=\left(3n-1\right)\left(3n+2\right)$
therefore, the series will be like $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(3n-1\right)\left(3n+2\right)}$
or, $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{9\left(n-\frac{1}{3}\right)\left(n+\frac{2}{3}\right)}$
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 $\sum _{n=1}^{\mathrm{\infty }}\left(\frac{1}{n+k}\right)\left(\frac{1}{n+k+1}\right)=\frac{1}{k+1}$
Applying these result to given series,
$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{9\left(n-\frac{1}{3}\right)\left(n+\frac{2}{3}\right)}=\frac{1}{9}\sum _{n=1}^{\mathrm{\infty }}\frac{1}{\left(n-\frac{1}{3}\right)\left(n+\frac{2}{3}\right)}=\frac{1}{9}×\frac{1}{\frac{2}{3}}=\frac{3}{18}=\frac{1}{6}$
Hence, the sum of the given convergent series is $\frac{1}{6}$
Jeffrey Jordon