The given series is

\(\sum_{n=1}^\infty\frac{1}{n-i^n}\)

Let us solve the given problem by the help of the Ratio test.

The Ratio test is,

For a series \(a_n,\text{ If }\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=L\)

Then, the series is convergent if L

the series is divergent if L>1

And if L=1, we have no result or we can say the test fails.

Here \(a_n=\frac{1}{n-i^n}\)

So, \(\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{n-i^n}{(n+1)-i^{n+1}}|(\frac{\infty}{\infty})\)

\(=\lim_{n\rightarrow\infty}|\frac{1-i^n\ln i}{1-i^{n+1}\ln i}|(\frac{\infty}{\infty})\)

\(=\lim_{n\rightarrow\infty}|\frac{i^n(\ln i)^2}{-i^{n+1}(\ln i)^2}|\)

\(=\frac1i\)

\(=\frac1i(\frac{i}{i})\)

=-i,(as \(i^2=-1\))

Therefore, L=-i

So, the given series is convergent.

ANSWER: The given series is convergent.

\(\sum_{n=1}^\infty\frac{1}{n-i^n}\)

Let us solve the given problem by the help of the Ratio test.

The Ratio test is,

For a series \(a_n,\text{ If }\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=L\)

Then, the series is convergent if L

the series is divergent if L>1

And if L=1, we have no result or we can say the test fails.

Here \(a_n=\frac{1}{n-i^n}\)

So, \(\lim_{n\rightarrow\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\rightarrow\infty}|\frac{n-i^n}{(n+1)-i^{n+1}}|(\frac{\infty}{\infty})\)

\(=\lim_{n\rightarrow\infty}|\frac{1-i^n\ln i}{1-i^{n+1}\ln i}|(\frac{\infty}{\infty})\)

\(=\lim_{n\rightarrow\infty}|\frac{i^n(\ln i)^2}{-i^{n+1}(\ln i)^2}|\)

\(=\frac1i\)

\(=\frac1i(\frac{i}{i})\)

=-i,(as \(i^2=-1\))

Therefore, L=-i

So, the given series is convergent.

ANSWER: The given series is convergent.