Does this convergence? Justify your answer..

An=(n^2+2^n)/((n^2)+2)

2 mark question from University past paper which is bugging me slightly.

I say that An is a sequence of non-negative integers and is not bounded, therefore, An diverges.

Is that enough of a justification or am I missing something here?

How do you know the terms are integers - I don't think they are!

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

How do you know the terms are integers - I don't think they are!

You basically need to be comparing the growth of the numerator with that of the denominator, which should be fairly easy to do since the n^2 terms are the same so you just need to say something about 2^n compared to 2.

It has been a while but I would have looked at

$\dfrac{1+\frac{2^n}{n^2}}{1+\frac{2}{n^2}}$

Since $\frac{2}{n^2}$ tends to 0 I would be considering$\frac{2^n}{n^2}$

Which, of course, does diverge

latex fail - not sure why

It has been a while but I would have looked at

$\dfrac{1+\frac{2^n}{n^2}}{1+\frac{2}{n^2}}$

Since $\frac{2}{n^2}$ tends to 0 I would be considering$\frac{2^n}{n^2}$

Which, of course, does diverge

latex fail - not sure why

One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

One more..

Summation of (n/n+1)

which series convergence test is recommended? I tried ratio which gave (n^2+2n)/(n^2+2n+1) which doesn't seem to help because this gives convergence to 1 (by diving through by n^2) which means the ratio test tells us nothing.

Apart from the issue of it not necessarily being an integer (take $n=1$), you should justify that it is unbounded (the whole integer / non-negative part is entirely unnecessary, since every convergent sequence of real numbers is bounded). Try the comparison test.



The null sequence test gives it immediately.

Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges

Thanks for the extra verification.

Last one for today:

Summation of an=n/(n^3-1) from n=2 to n--->infinity

By comparison test, we get that an is less than 1/n^2. Since 1/n^2 converges to 0 as n--->infiniry, we conclude that an tends to 0 as n---> infinity. since 0<1, we conclude that summation of an converges