AP Calculus BC

Does the Sum of n^2/(n^2+1) Converge? No

The series diverges. The terms approach one rather than zero, so the nth term test settles it in a single line. Terms that shrink toward a nonzero number are still being added forever, and the total runs away.

\sum_{n=1}^{\infty}\frac{n^{2}}{n^{2}+1}

Diverges

Settled by the nth term test for divergence.

One limit finishes it

\lim_{n \to \infty}\frac{n^{2}}{n^{2}+1} = \lim_{n \to \infty}\frac{1}{1 + 1/n^{2}} = 1 \ne 0

Since the terms do not tend to 0, the series diverges. No further test is needed, and running one would be wasted effort.

The intuition is worth holding onto: past a certain point every term is at least $0.99$ , so adding a hundred more terms adds at least 99 to the total. There is no ceiling.

The trap in the shape

The terms are all less than 1 and they are increasing toward it, which can read as settling down. Bounded terms are not shrinking terms, and only shrinking to ZERO is relevant.

Compare with $\sum \frac{1}{n^{2}+1}$ , which converges. Same denominator, completely different verdict, because there the numerator is constant while the denominator grows. Always compute the limit of the terms before pattern matching on the denominator.

Not sure which test a series wants?

The Convergence Test Chooser walks the decision in order: nth term first, then geometric and p-series pattern matching, then alternating structure, then the ratio test, and finally the comparison family.

Frequently asked questions

Why does the nth term test come first?

It costs one limit and often ends the problem. If the terms fail to tend to 0 you are finished; if they do tend to 0 you have lost almost nothing and move on.

Can this series converge if I start it later?

No. Dropping finitely many terms never changes convergence, and the terms still approach 1 no matter where you start.