"Consider the following series. ∑n=1∞n+4n2 The series is..." was answered by students like you

iohanetc

Answered question

2021-03-06

Consider the following series.

\sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}}

The series is equivalent to the sum of two p-series. Find the value of p for each series.
Determine whether the series is convergent or divergent.

Arham Warner

Skilled2021-03-07Added 102 answers

Given the series:

\sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}}

Rewriting the given series:

\Rightarrow \sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}} = \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^{2}} + \frac{4}{n^{2}}

\Rightarrow \sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}} = \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^{2}} + \sum_{n = 1}^{\infty} \frac{4}{n^{2}}

\Rightarrow \sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}} = \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^{2 - \frac{1}{2}}} + 4 \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

\Rightarrow \sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}} = \sum_{n = 1}^{\infty} \frac{\sqrt{n}}{n^{\frac{3}{2}}} + 4 \sum_{n = 1}^{\infty} \frac{1}{n^{2}}

Hence we get sum of two p series with:

p_{1} = \frac{3}{2}

p_{2} = 2

Since both

p_{1} > 1, p_{2} > 1

, both the individual p series converge.
The given series is sum of two converging series, hence the given series

\sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}}

converges.
Final Answer:

p_{1} = \frac{3}{2}

p_{2} = 2

\sum_{n = 1}^{\infty} \frac{\sqrt{n} + 4}{n^{2}}

converges

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