What is the sum from n=1 to infinity of (99^n(n^5+2))/(n^210^{2n)} ?

The sum from n=1 to infinity of (99^n(n^5+2))/(n^210^{2n)} is converges

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sum from n=1 to infinity of (99^n(n^5+2))/(n^210^{2n)}

n = 1 \sum \infty \frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} 1 0 ^{2 n}}

converges

Solution steps

n = 1 \sum \infty \frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} \cdot 1 0 ^{2 n}}

Expand

\frac{9 9 ^{n} ( n ^{5} + 2 )}{n ^{2} \cdot 1 0 ^{2 n}} : \frac{9 9 ^{n} n ^{3}}{10 0 ^{n}} + \frac{2 \cdot 9 9 ^{n}}{10 0 ^{n} n ^{2}}

= n = 1 \sum \infty \frac{9 9 ^{n} n ^{3}}{10 0 ^{n}} + \frac{2 \cdot 9 9 ^{n}}{10 0 ^{n} n ^{2}}

Apply the Sum Rule:

\sum a_{n} + b_{n} = \sum a_{n} + \sum b_{n}

= n = 1 \sum \infty \frac{9 9 ^{n} n ^{3}}{10 0 ^{n}} + n = 1 \sum \infty \frac{2 \cdot 9 9 ^{n}}{10 0 ^{n} n ^{2}}

n = 1 \sum \infty \frac{9 9 ^{n} n ^{3}}{10 0 ^{n}} =

converges

n = 1 \sum \infty \frac{2 \cdot 9 9 ^{n}}{10 0 ^{n} n ^{2}} =

converges

= converges + converges

= converges

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What is the sum from n=1 to infinity of (99^n(n^5+2))/(n^210^{2n)} ?
The sum from n=1 to infinity of (99^n(n^5+2))/(n^210^{2n)} is converges