Step 1

Calculation:

\(\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{\frac{{{10}}}{{{k}^{{{2}}}+{9}}}}\)

\(\displaystyle={10}{\sum_{{{k}={0}}}^{{\infty}}}{\frac{{{1}}}{{{k}^{{{2}}}+{9}}}}\)

Applying series comparison test

\(\displaystyle{k}^{{{2}}}+{9}{>}{k}^{{{2}}}\)

\(\displaystyle{\frac{{{1}}}{{{k}^{{{2}}}+{9}}}}{<}{\frac{{{1}}}{{{k}^{{{2}}}}}}\)

\(\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{\frac{{{1}}}{{{k}^{{{2}}}}}}\Rightarrow\) convergent by p series test.

Hence by comparison test

\(\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{\frac{{{1}}}{{{k}^{{{2}}}+{9}}}}\) is convergent.

Also \(\displaystyle{\sum_{{{k}={0}}}^{{\infty}}}{\frac{{{10}}}{{{k}^{{{2}}}+{9}}}}\) is convergent.