Question

5 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!

Answer 1

Yes, this series converges. We can check using the integral test; we have

$\displaystyle\frac1{25}+\frac1{36}+\frac1{49}+\cdots=\sum_{n=5}^\infty\frac1{n^2}$

and

$\displaystyle\sum_{n=5}^\infty\frac1{n^2}\le\int_5^\infty\frac{\mathrm dx}{x^2}=\frac15$

Answer 2

Answer:

Option A is correct.

True.

The series: $\frac{1}{25}+\frac{1}{36} +\frac{1}{49} +....$ is convergent

Step-by-step explanation:

Comparison Test:

Let $0\leq a_n\leq b_n$ for all n.

If $\sum_{n=1}^{\infty} b_n$ converges, then  $\sum_{n=1}^{\infty} a_n$ converges.

If  $\sum_{n=1}^{\infty} b_n$ diverges, then  $\sum_{n=1}^{\infty} a_n$ diverges.

Given the series: $\frac{1}{25}+\frac{1}{36} +\frac{1}{49} +....$

then;

$a_n = \sum_{n=1}^{\infty}\frac{1}{(n+4)^2}$

$\frac{1}{(n+4)^2} \leq \frac{1}{n^2}$

By comparison test:

$b_n = \frac{1}{n^2}$

P-series test:

$\sum_{n=1}^{\infty} \frac{1}{n^p}$ where p> 0

If p>1 then the series converges and if 0<p< 1, then the series diverges.

By using p-test series in series $b_n$

then;

$b_n = \sum_{n=1}^{\infty} \frac{1}{n^2}$ is a p-series, with p> 1, it converges.

Comparing the above series with $b_n = \frac{1}{n^2}$, we can conclude that $a_n = \sum_{n=1}^{\infty}\frac{1}{(n+4)^2}$ also converges and $\frac{1}{(n+4)^2} \leq \frac{1}{n^2}$

Therefore, the given series is convergent.