Answer to Question #174298 in Calculus for Raj

Question

1:Examine the series for convergence for summation of ((-1)^(n-1) sin(nx))/n^3

2: Test for convergence of summation  (x^n)/(2n)!

3: Examine the following series for absolute convergence of x-(x^2/2)+(x^3/3)-(x^4/4)+......

4:Test convergence of series 

  1+(1*2^2)/(1*3*5)+(1*2^2*3^2)

5: Test convergence of summation (x^n)/n!

6:Examine the following series for absolute convergence of 1+(x/2)+(x^2/3^2)+(x^3/4^3)+......(x>0)

7: Examine convergence of summation 1/root n

Answer 1

$\begin{array}{l} \text { 7) } \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}, \text { by using integral test, } \\ \int_{1}^{\infty} \frac{1}{\sqrt{n}} d n=(\ln \sqrt{n}) /_{1}^{\infty}=\ln (\infty)-\ln 1=\infty \\ \text { So the series diverges } \end{array}$ $\begin{array}{l} \text { 6) } 1+\frac{x}{2}+\frac{x^{2}}{3^{2}}+\frac{x^{3}}{4^{3}}+\ldots(x>0)= \\ =\sum_{n=0}^{\infty} \frac{x^{n}}{(n+1)^{n}} \text { by using Root test }\left(a_{k}\right)^{1 / k}<1 \Rightarrow \text { converent } \\ \left(\frac{x^{n}}{(n+1)^{n}}\right)^{1 / n}=\frac{x}{n+1}<1, \text { if } x<n+1 \text { , then convergent } \end{array}$ $\begin{array}{l} \text { 5) } \sum_{n=0}^{\infty} \frac{x^{n}}{n !} \text { by using Ratio test } \frac{a_{k+1}}{a_{k}}<1 \text { , converge } \\ \frac{x^{n+1}}{(n+1) !} \cdot \frac{n !}{x^{n}}=\frac{x \cdot x^{n}}{n !(n+1)} \cdot \frac{n !}{x^{n}} \Rightarrow \frac{x}{n+1}<1 \\ \text { if } x<n+1, \text { the series is convereent } \end{array}$ $\begin{array}{l} \text { 3) } x-\left(\frac{x^{2}}{2}\right)+\left(\frac{x^{3}}{3}\right)-\left(\frac{x^{4}}{4}\right)+\ldots \Rightarrow \sum_{n=1}^{\infty}(-1)^{n+1} \cdot \frac{x^{n}}{n} \\ \text { by using alternating test if } \frac{x^{n}}{n} \text { is monotone, } \\ \text { and if } \lim _{n \rightarrow \infty} \frac{x^{n}}{n}=0 \text { , then the series converges, } \\ \text { if }-1<x \leqslant 1 \quad \rightarrow \text { converpent } \end{array}$ $\begin{array}{l} \text { 2) } \sum_{n=0}^{\infty} \frac{x^{n}}{(2 n) !} \text { by wing Ratio test } \frac{a_{k+1}}{a_{k}}<1 \text { , converges. } \\ \frac{x^{n+1}}{(2 n+1) !} \cdot \frac{2 n !}{x^{n}}=\frac{x \cdot x^{n}}{2 n !(2 n+1)} \cdot \frac{2 n !}{x^{n}}=\frac{x}{2 n+1} \\ \text { if } \frac{x}{2 n+1}<1, \text { converges } \Rightarrow x<2 n+1 \end{array}$ $\begin{array}{l} \text { 1) } \sum_{n=1}^{\infty} \frac{(-1)^{(n-1)} \cdot \sin (n x)}{n^{3}} \quad \text { if } \frac{\sin (n x)}{n^{3}} \rightarrow \text { momotone } \\ \text { for any } x \in N \quad \lim _{n \rightarrow \infty} \frac{\sin (n x)}{n^{3}}=0 \\ \text { So series converges }\\ \text {4) Conditions of Question 4 are not full } \end{array}$