Question

Please help me solve these series qns! T^T

Answer 1

I'm not so sure about the first two, but for the third:
\[\sum_{n=1}^\infty \frac{2+(-1)^n}{n\sqrt n}=\sum_{n=1}^\infty \frac{2}{n\sqrt n}+\sum_{n=1}^\infty \frac{(-1)^n}{n\sqrt n}\]
The first series clearly converges, since the degree in the denominator is greater than 1. Apply the alternating series test to the second series: show that the non-alternating part is decreasing, and that it approaches 0 as \(n\to\infty\). Clearly, \(\dfrac{1}{n\sqrt n}\) is a decreasing sequence, and \(\displaystyle\lim_{n\to\infty}\frac{1}{n\sqrt n}=0\), so the second series also converges. Since both series converge, the original series must also be convergent.

Answer 2

@SithsAndGiggles Thank you! that helped alot! (: now the first 2 qns...

Answer 3

First question:

Info you need to solve (integral test):
Limit as b approaches infinity of the integral of 1/x^5dx from n to b
-1/4limit as b approaches infinity of the integral of 1/x^4dx from n to b
-1/4limit as b approaches infinity of (1/b^4-1/n^4)
1/(4n^4)

The integral test shows that your integral from n to infinity of 1/x^5 is an upper bound for the sum from n to infinity of 1/n^5. You want to find n big enough so that the sum of this tail, the remaining part of the series that you'll be throwing out, is less than 0.0005, which is 1/2000;

So, you want to solve the inequality 1/(4 n^4) < 1/2000.

Cross multiplying gives 2000 < 4 n^4; divide by 4: 500 < n^4; take 4th roots: 4.8 < n

a(n)=1/n⁵
a(1)=1
a(2)=1/2⁵=1/32=0.03125
a(3)=1/3⁵=1/243≈0.004115
a(4)=1/4⁵=1/1024≈0.00097656

∞
∑ 1/n^5 ≈ a(1)+a(2)+a(3)+a(4) ≈ 1.03634 ≈ 1.036
n=1

Answer 4

@swaim hi! thanks for the solution! :D however i dont get the part how u get the 0.0005, i mean why do we have to solve the inequality.. ( integral < 1? ) why?