# Integral Test/Comparison Test

• Jan 27th 2009, 10:51 AM
Integral Test/Comparison Test
infinity over Sigma n=1 n^2/n^3 + 1 Use the integral test to see if the infinite series converges.

infinity over Sigma n=2 n^2+1/n^3.5-2 Use either the Comparison Test or the Limit Comparison Test to see if the infinite series converges.
• Jan 27th 2009, 10:58 AM
Krizalid
Quote:

infinity over Sigma n=1 n^2/n^3 + 1 Use the integral test to see if the infinite series converges.

Put $\displaystyle f(x)=\frac{{{x}^{3}}}{{{x}^{2}}+1},$ then $\displaystyle f$ is a positive, continuous and decreasing function for $\displaystyle x\ge2,$ hence, the integral test applies and the series will converge or diverge if $\displaystyle \int_{2}^{\infty }{\frac{{{x}^{3}}}{{{x}^{2}}+1}\,dx}$ does.

Since $\displaystyle \frac{{{x}^{3}}}{{{x}^{2}}+1}\ge \frac{{{x}^{3}}}{{{x}^{2}}+{{x}^{2}}}=\frac{1}{2}x ,$ then the integral diverges, so does the series. (Actually, one can bound immediately the general term of the series rather than applying the integral test.)

-----

As for your second question, compare the series with $\displaystyle \sum\limits_{n=2}^{\infty }{\frac{1}{{{n}^{1,5}}}}$ which is a convergent $\displaystyle p-$series with $\displaystyle p=1,5>1.$

So, put $\displaystyle {{a}_{n}}=\frac{{{n}^{2}}+1}{{{n}^{3,5}}-2}$ and $\displaystyle {{b}_{n}}=\frac{1}{{{n}^{1,5}}},$ and observe that $\displaystyle \underset{n\to \infty }{\mathop{\lim }}\,\frac{{{a}_{n}}}{{{b}_{n}}}=1,$ so your series converges.