"Does the attached series converge or diverge? Note:..." was answered by students like you

Clifland

Answered question

2021-01-04

Does the attached series converge or diverge? Note: it is a complex series and not a real series. Ive

Answer & Explanation

Laaibah Pitt

Skilled2021-01-05Added 98 answers

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 $a_{n}, If lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = L$
Then, the series is convergent if $L < 1$
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, $lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = lim_{n \to \infty} | \frac{n - i^{n}}{(n + 1) - i^{n + 1}} | (\frac{\infty}{\infty})$
$= lim_{n \to \infty} | \frac{1 - i^{n} \ln i}{1 - i^{n + 1} \ln i} | (\frac{\infty}{\infty})$
$= lim_{n \to \infty} | \frac{i^{n} (\ln i)^{2}}{- i^{n + 1} (\ln i)^{2}} |$
$= \frac{1}{i}$
$= \frac{1}{i} (\frac{i}{i})$
=-i,(as $i^{2} = - 1$ )
Therefore, $L = - i < 1$
So, the given series is convergent.
ANSWER: The given series is convergent.

Jeffrey Jordon

Expert2021-12-16Added 2605 answers

Answer is given below (on video)

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