Question

Is this series convergent or divergent ?

Answer 1

\[\large{\sum_{n=1}^{\infty}[(n^3 + 1)^{(1/3)} - n]}\]

Answer 2

@ganeshie8

Answer 3

I have tried using limit comparison test. The test is inconclusive.

Not sure if ratio test is going to help.

Okay let me see:

\[\large{a_n = (n^3 + 1)^{(1/3)}-n}\]

\[\large{b_n = \cfrac{1}{n^2 + n +1}>\cfrac{1}{n^2} \rm{which \ is \ convergent}}\]

Thus, by direct comparison test, \(\large{b_n}\) is also convergent.

Now:

\[\large{\lim_{n \rightarrow \infty} \cfrac{a_n}{b_n} = 1 > 0}\]

Thus, \(a_n\) and \(b_n) both converge.

Hurray !!! Got it :)

Answer 4

Okay forget it... its wrong :(

Answer 5

\(\color{blue}{\text{Originally Posted by}}\) @vishweshshrimali5
I have tried using limit comparison test. The test is inconclusive.

Not sure if ratio test is going to help.

Okay let me see:

\[\large{a_n = (n^3 + 1)^{(1/3)}-n}\]

\[\color{red}{\large{b_n = \cfrac{1}{n^2 + n +1}>\cfrac{1}{n^2} \rm{which \ is \ convergent}}}\]

Thus, by direct comparison test, \(\large{b_n}\) is also convergent.

Now:

\[\large{\lim_{n \rightarrow \infty} \cfrac{a_n}{b_n} = 1 > 0}\]

Thus, \(a_n\) and \(b_n\) both converge.

Hurray !!! Got it :)
\(\color{blue}{\text{End of Quote}}\)


That step is wrong.

But if I replace it by:

\[\color{green}{\large{b_n = \cfrac{1}{n^2 + n +1}<\cfrac{1}{(n+1)^2} \rm{which \ is \ convergent}}}\]

Then it is right :)

Answer 6

I can also prove that \(b_n\) is convergent using limit comparison test:

\[\large{a'_n = \cfrac{1}{n^2 + n + 1}}\]

\[\large{b'_n = \cfrac{1}{n^2} \rm{which \ is \ convergent}}\]

Now:

\[\large{\lim_{n \rightarrow \infty}\cfrac{a'_n}{b'_n}}\]

\[\large{=\lim_{n \rightarrow \infty} \cfrac{1}{1+ \cfrac{1}{n} + \cfrac{1}{n^2}}}\]

\[\large{= 1 > 0}\]

Thus, \(a'_n\) must be convergent.

This also means that \(a_n\) is convergent.

QED :)

Answer 7

Worked !!

Can someone check this please ?

Answer 8

\[\frac{(n^3+1)^\frac{1}{3}-n}{1} \cdot \frac{(n^3+1)^\frac{2}{3}+(n^3+1)^\frac{1}{3}n+n^2}{(n^3+1)^\frac{2}{3}+(n^3+1)^\frac{1}{3}n+n^2}\\ =\frac{1}{(n^3+1)^\frac{2}{3}+(n^3+1)^\frac{1}{3}n+n^2} \le \frac{1}{n^2+n(n)+n^2}=\frac{1}{3n^2}\]
I think it would have been good to compare these right?

Answer 9

\[\int\limits_{1}^{\infty}\frac{1}{3n^2} dn =\int\limits_{1}^{\infty}\frac{1}{3}n^{-2} dn =\lim_{a \rightarrow \infty} (\frac{-1}{3a}+\frac{1}{3})=0+\frac{1}{3}\]
By integral test 1/(3n^2) converges
so so does the series you mentioned by comparison.

Answer 10

Great work @myininaya

Thanks a lot.

An alternate method is always good :)