Question

Test each following to convergance
∑(n=1,∞)(√(n^3 +5n -1)/(n^2 - sin(n^3))


2. the sequence { (n^2 /(3*n +1 ))sin ((\pi[/tex])/(3*n ) } for all n≥1


pi =3.14.

Answer 1

1. Compare the series to the divergent p-series,

$\displaystyle\sum_{n=1}^\infty\frac1{\sqrt n}$

Use the limit comparison test: we have

$\displaystyle\lim_{n\to\infty}\frac{\frac{\sqrt{n^3+5n-1}}{n^2-\sin(n^3)}}{\frac1{\sqrt n}}=1$

so the given series also diverges.

Breakdown for the limit: $\sin(n^3)$ oscillates between -1 and 1, so its contribution to the denominator $n^2-\sin(n^3)$ is neglible and we can replace it with $n^2$. In the square root in the numerator, the $n^3$ term dominates, so $\sqrt{n^3+5n-1}\approx\sqrt{n^3}=n^{3/2}$. So the given summand is approximated by

$\dfrac{n^{3/2}}{n^2}=\dfrac1{n^{1/2}}=\dfrac1{\sqrt n}$

2. Let $m=\frac\pi{3n}$, so that $m\to0$ as $n\to\infty$. Then

$\displaystyle\lim_{n\to\infty}\frac{n^2}{3n+1}\sin\left(\dfrac\pi{3n}\right)=\lim_{m\to0}\frac{\frac{\pi^2}{9m^2}}{\frac\pi m+1}\sin m$

Recall that $\frac{\sin x}x\to1$ as $x\to0$. Rewrite the limit as

$\displaystyle\lim_{m\to0}\frac{\frac{\pi^2}9}{m+\pi}\frac{\sin m}m$

Then $\frac{\sin m}m\to1$, and we're left with

$\displaystyle\lim_{m\to0}\frac{\frac{\pi^2}9}{m+\pi}$

which is continuous at $m=0$, giving a limit of $\frac\pi9$.