Respuesta :

LammettHash

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

[tex]\displaystyle\frac1{25}+\frac1{36}+\frac1{49}+\cdots=\sum_{n=5}^\infty\frac1{n^2}[/tex]

and

[tex]\displaystyle\sum_{n=5}^\infty\frac1{n^2}\le\int_5^\infty\frac{\mathrm dx}{x^2}=\frac15[/tex]

Answer:

Option A is correct.

True.

The series: [tex]\frac{1}{25}+\frac{1}{36} +\frac{1}{49} +....[/tex] is convergent

Step-by-step explanation:

Comparison Test:

Let [tex]0\leq a_n\leq b_n[/tex] for all n.

If [tex]\sum_{n=1}^{\infty} b_n[/tex] converges, then  [tex]\sum_{n=1}^{\infty} a_n[/tex] converges.

If  [tex]\sum_{n=1}^{\infty} b_n[/tex] diverges, then  [tex]\sum_{n=1}^{\infty} a_n[/tex] diverges.

Given the series: [tex]\frac{1}{25}+\frac{1}{36} +\frac{1}{49} +....[/tex]

then;

[tex]a_n = \sum_{n=1}^{\infty}\frac{1}{(n+4)^2}[/tex]

[tex]\frac{1}{(n+4)^2} \leq \frac{1}{n^2}[/tex]

By comparison test:

[tex]b_n = \frac{1}{n^2}[/tex]

P-series test:

[tex]\sum_{n=1}^{\infty} \frac{1}{n^p}[/tex] 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 [tex]b_n[/tex]

then;

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

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

Therefore, the given series is convergent.

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