Answer to Question #252826 in Calculus for Sayem

Question

Determine whether the following series converge, converge absolutely, converge conditionally, or diverge.

1. $\displaystyle\sum_{k=1}^ ∞$ $√k/k^2+k+1$
2. $\displaystyle\sum_{k=1}^ ∞$ $k!/(2^k*k^4)$
3. $\displaystyle\sum_{k=1}^ ∞$ $[k+(1/k)]^2/(k^2+1)$^3/2

( As I speak English as a Second language, Can you kindly explain a little bit so that I can do other similar problems by myself and any resources where I should study further theories of Sequence and Series) Thank you.

Answer 1

1. Consider the given series: $\sum_{k=1}^{\infty }\frac{\sqrt{k}}{k^2+k+1}$

Use Limit Comparison Test, to test the convergence of the give series.

Let $a_{k}=\frac{\sqrt{k}}{k^2+k+1}$

Take $b_{k}=\frac{\sqrt{k}}{k^2}=\frac{1}{k^{\frac{3}{2}}}$

Using Limit Comparison Test, we have

$\lim_{k\rightarrow \infty }\frac{a_{k}}{b_k}=\lim_{k\rightarrow \infty }\frac{\sqrt{k}}{k^2+k+1}\times \frac{k\sqrt{k}}{1}=\lim_{k\rightarrow \infty }\frac{k^2}{k^2+k+1}=1\neq 0$

So, Limit Comparison Test can be applied

And the series $\sum_{k=1}^{\infty }\frac{1}{k^{3/2}}$ is convergent by using $p-$ series $(p=\frac{3}{2}>1)$

So, By Limit Comparison Test, the given series $\sum_{k=1}^{\infty }\frac{\sqrt{k}}{k^2+k+1}$ is also convergent.

Observe that the given series is a series of positive terms.

Therefore, the given series is absolutely convergent.

2.Consider the series: $\sum_{k=1}^{\infty }\frac{k!}{2^{k}\times k^{4}}$

Use Ratio Test to test the convergence of the given series.

Let $a_{k}=\frac{k!}{2^{k}\times k^{4}}$

Then, $a_{k+1}=\frac{(k+1)!}{2^{k+1}\times (k+1)^{4}}$

By using Ratio Test, we have $L=\lim_{k\rightarrow \infty }\frac{a_{k+1}}{a_{k}}$

Then, we get $L=\lim_{k\rightarrow \infty }\frac{(k+1)!}{2^{k+1}\times (k+1)^{4}}\times \frac{2^{k}\times k^{4}}{k!}$

By simplification, we get

$L=\lim_{k\rightarrow \infty }\frac{k+1}{2}\times (\frac{k}{k+1})^{4}=\frac{1}{2}\times (1)\times (\infty )=\infty >1$

So, the given series is divergent by Ratio Test.

Therefore, the given series is divergent by Ratio Test.

3.Consider the series: $\sum_{k=1}^{\infty }\frac{[k+(\frac{1}{k})]^2}{(k^{2}+1)^{3/2}}$

Let $a_{k}=\frac{(k^2+1)^2}{k^2(k^2+1)^{3/2}}$

Use Limit Comparison Test to test the convergence of the given series

Take $b_{k}=\frac{k^4}{k^5}=\frac{1}{k}$

By using Limit Comparison Test, we have

$L=\lim_{k\rightarrow \infty }\frac{a_{k}}{b_{k}}=\lim_{k\rightarrow \infty }\frac{(k^2+1)^2}{k^2(k^2+1)^{3/2}}\times \frac{k}{1}$

Then, we get $L=1\neq 0$

So, Limit Comparison Test can be applied.

Now $\sum_{k=1}^{\infty }b_k=\sum_{k=1}^{\infty }\frac{1}{k}$ is divergent by using $p-$ series $(p=1)$

Therefore, by Limit Comparison Test, the given series $\sum_{k=1}^{\infty }\frac{[k+(\frac{1}{k})]^2}{(k^{2}+1)^{3/2}}$ is also divergent.