AP Calculus BC

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

Q: Does the sum of n^2/(2n^2+1) converge?

No. Its terms tend to \frac{1}{2}, so it diverges by the nth term test.

The sum of n squared over 2n squared plus 1 diverges. The terms approach one half, not 0, so the nth term test proves divergence immediately. Any series whose terms settle at a nonzero value must diverge.

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

Diverges

Settled by the nth term test for divergence.

Equal degrees give a nonzero limit

Dividing through by $n^{2}$ makes the limit obvious.

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

The terms settle near $\frac{1}{2}$ , so after enough terms you are adding about $\frac{1}{2}$ every time and the total grows without bound.

The mistakes students make

Cancelling $n^{2}$ from the numerator and the $2n^{2}$ only, ignoring the $+1$ . Divide EVERY term.
Assuming a fraction whose parts both grow must tend to $0$ .
Skipping to the comparison tests. The nth term limit is nonzero, so nothing else is needed.

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

Does the sum of n^2/(2n^2+1) converge?

No. Its terms tend to $\frac{1}{2}$ , so it diverges by the nth term test.

Why is the limit not 0?

Numerator and denominator have the same degree, so the limit is the ratio of leading coefficients, $\frac{1}{2}$ .