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

5 (Picture) CONVERGENT AND DIVERGENT SERIES PLEASE HELP!!-example-1
User Zajn
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2 Answers

4 votes

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(\mathrm dx)/(x^2)=\frac15

User Pankhuri Agarwal
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8.1k points
4 votes

Answer:

Option A is correct.

True.

The series:
(1)/(25)+(1)/(36) +(1)/(49) +.... is convergent

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:
(1)/(25)+(1)/(36) +(1)/(49) +....

then;


a_n = \sum_(n=1)^(\infty)(1)/((n+4)^2)


(1)/((n+4)^2) \leq (1)/(n^2)

By comparison test:


b_n = (1)/(n^2)

P-series test:


\sum_(n=1)^(\infty) (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) (1)/(n^2) is a p-series, with p> 1, it converges.

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

Therefore, the given series is convergent.

User Ivan Lesko
by
8.5k points

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