Question

Determine whether the following series converges or diverges.

Answer 1

\[\sum_{n=1}^{\infty} n/\sqrt{n^8 +3}\]

Answer 2

I used the Ratio test and this is what I got but I'm having difficulty simplifying it.

Answer 3

try comparison test instead

Answer 4

\[\left| (n+1)/\sqrt{(n+1)^8 +3} * \sqrt{(n^8 +3)}/n \right|\]

Answer 5

Okay, I will try that.

Answer 6

Notice that \(\large \sqrt{n^8+3}\gt \sqrt{n^8} \implies \dfrac{1}{\sqrt{n^8+3}}\lt \frac{1}{\sqrt{n^8}}\)

Answer 7

Oh and if you simplify 1/(n^8)^(1/2) you'll get 1/n^4 which you can conclude is convergent by the p series right?

Answer 8

Yes that should work, don't forget the "n" in the numerator though

Answer 9

\[\large {\sqrt{n^8+3}\gt \sqrt{n^8} \implies \dfrac{1}{\sqrt{n^8+3}}\lt \frac{1}{\sqrt{n^8}} \\~\\\implies \dfrac{n}{\sqrt{n^8+3}}\lt \frac{n}{\sqrt{n^8}} =\frac{1}{n^3}}\]



Since \(\sum\frac{1}{n^3}\) converges by p-series test,
the series with smaller terms, \(\sum\frac{n}{\sqrt{n^8+3}}\), converges by comparison test.

Answer 10

Okay, so you get n/n^4 which just simplifies to 1/n^3 and we can conclude divergence based on p-series test. Thanks :)