Does the series $\infty \sum n = 1 \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n$ converge or diverge?

Question

Does the series $\infty \sum n = 1 \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n$ converge or diverge?

1 Answer

Impact of this question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-18T16:35:50

See below.

Explanation:

Considering

$a_{n} = \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $a_n = (5n)^(3n)/(5^n+3)^n$

we have

$a_{n} < \frac{{(5 n)}^{3 n}}{{(5^{n})}^{n}}$ $a_n < (5n)^(3n)/(5^n)^n$

now we will compare $a_{n}$ $a_n$ with $\frac{1}{n^{2}}$ $1/n^2$ that we know to converge.

Taking $ln$ $ln$

${ln a}_{n} < (3 n) ln (5 + ln n) - n^{2} ln 5$ $ln a_n < (3n)ln(5+ln n)-n^2 ln 5$ and as we can easily verify exists a $n_{0}$ $n_0$ such that $n > n_{0} \Rightarrow {ln a}_{n} < ln (\frac{1}{n^{2}})$ $n > n_0 rArr ln a_n < ln (1/n^2)$

As we know, $ln$ $ln$ is a strictly increasing function so considering $n > n_{0}$ $n > n_0$

${ln a}_{n} < ln (\frac{1}{n^{2}}) \Rightarrow a_{n} < \frac{1}{n^{2}}$ $lna_n < ln (1/n^2) rArr a_n < 1/n^2$

hence the series

$\infty \sum n = 1 \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n$

converges.

Science

Math

Humanities

... and beyond

Does the series $\infty \sum n = 1 \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n$ converge or diverge?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Does the series sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n∞∑n=1(5n)3n(5n+3)n sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n converge or diverge?

1 Answer

Explanation:

Related questions

Impact of this question

Does the series $\infty \sum n = 1 \frac{{(5 n)}^{3 n}}{{(5^{n} + 3)}^{n}}$ $sum_{n=1}^oo (5n)^(3n)/(5^n+3)^n$ converge or diverge?