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prisoha [69]
3 years ago
11

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

Mathematics
2 answers:
ella [17]3 years ago
6 0

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

mojhsa [17]3 years ago
6 0

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.

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