Answer to Question #260595 in Real Analysis for saduni

Question

Find the convergence of the following series,

1. ∑^∞ _𝑛=1(√𝑛+1−√𝑛)/ 𝑛  
2. ∑^∞ _𝑛=1 cos(1/n)
3. ∑^∞ _𝑛=1 (sin n)/n
4. ∑^∞ _𝑛=1 ((𝑛!) ²)/((2𝑛)!)
5. ∑^∞ _𝑛=1 (n/n+1)n²

Answer 1

1.

$$\displaystyle\sum_{n=1}^{\infin}\dfrac{\sqrt{n+1}-\sqrt{n}}{n}$$

$$=\displaystyle\sum_{n=1}^{\infin}\dfrac{\sqrt{n+1}-\sqrt{n}}{n}(\dfrac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}})$$

$$=\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n(\sqrt{n+1}+\sqrt{n})}$$

$$=\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n^{3/2}(\sqrt{1+1/n}+1)}$$

The p-series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n^{3/2}}$ is convergent (p=3/2).

$$\dfrac{1}{n^{3/2}(\sqrt{1+1/n}+1)}<\dfrac{1}{2n^{3/2}}, n\geq1$$

Then the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{1}{n^{3/2}(\sqrt{1+1/n}+1)}$ converges by the Comparison Test.

Therefore the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{\sqrt{n+1}-\sqrt{n}}{n}$ converges by the Comparison Test.

2.

$$\displaystyle\sum_{n=1}^{\infin}\cos(\dfrac{1}{n})$$

$$\lim\limits_{n\to\infin}\cos(\dfrac{1}{n})=\cos(0)=1\not=0$$

Therefore the series $\displaystyle\sum_{n=1}^{\infin}\cos(\dfrac{1}{n})$ diverges by the Test for Divergence.

3.

$$\displaystyle\sum_{n=1}^{\infin}\dfrac{\sin(n)}{n}$$

Use the Dirichlet's Test

$$2\sin(1)\displaystyle\sum_{n=1}^{m}\sin(n)=\displaystyle\sum_{n=1}^{m}(\cos(n-1)-\cos(n+1))$$

$$=(\cos(0)-\cos(2))+(\cos(1)-\cos(3))$$

$$+(\cos(2)-\cos(4))+...+(\cos(m-2)-\cos(m))$$

$$+(\cos(m-1)-\cos(m+1))$$

$$=1+\cos(1)-\cos(m)-\cos(m+1)$$

Then

$$\displaystyle\sum_{n=1}^{m}\sin(n)=\dfrac{1+\cos(1)-\cos(m)-\cos(m+1)}{2\sin(1)}$$

$$\bigg|\displaystyle\sum_{n=1}^{m}\sin(n)\bigg|=\dfrac{|1+\cos(1)-\cos(m)-\cos(m+1)|}{2\sin(1)}$$

$$\leq \dfrac{1+1+1+1}{2\sin(1)}=\dfrac{2}{\sin(1)}=M$$

Let $a_n=\dfrac{1}{n}.$ Then

$$a_{n+1}=\dfrac{1}{n+1}<\dfrac{1}{n}=a_n, n\geq1$$ $$\lim\limits_{n\to\infin}a_n=\lim\limits_{n\to\infin}\dfrac{1}{n}=0$$

Therefore the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{\sin(n)}{n}$ converges by the Dirichlet's Test.

4.

$$\displaystyle\sum_{n=1}^{\infin}\dfrac{(n!)^2}{(2n)!}$$

Apply the Ratio Test.

$$\lim\limits_{n\to\infin}\dfrac{\dfrac{((n+1)!)^2}{(2(n+1))!}}{\dfrac{(n!)^2}{(2n)!}}=\lim\limits_{n\to\infin}\dfrac{(n+1)^2}{(2n+1)(2n+2)}=\dfrac{1}{4}<1$$

Therefore the series $\displaystyle\sum_{n=1}^{\infin}\dfrac{(n!)^2}{(2n)!}$ converges by the RatioTest.

5.

$$\displaystyle\sum_{n=1}^{\infin}(\dfrac{n}{n+1})^{n^2}$$

Use the Root Test.

$$\lim\limits_{n\to\infin}\sqrt[n]{(\dfrac{n}{n+1})^{n^2}}=\lim\limits_{n\to\infin}(1-\dfrac{1}{n} )^n=e^{-1}=\dfrac{1}{e}<1$$

Therefore the series $\displaystyle\sum_{n=1}^{\infin}(\dfrac{n}{n+1})^{n^2}$ converges by the Root Test.