Question

Q3: determine wither the following series are converge or diverge In(n? +3) n!.inn a) Σ=1 n=2 (n+3) n (n+2)! (12 point) c) Σ=2 8 b) Σ=1 3+ (3)an d) Σ-1 In(n) 3 (m)2

Answer 1

a) $\frac{\ln(n^2+3)}{n+3}\ge \frac{1}{n+3}$ which is divergent so this diverges (as sum of reciprocals diverges)

b) $\frac{n!\ln(n)}{n(n+2)!}\le \frac{\ln(n)}{n(n+1)(n+2)}\le \frac{\ln(n)}{n^3}\le\frac{1}{n^2}$ as . The last series converges so this series must also converge

In(n) < n

c) $\frac{8}{\left(3+\frac{1}{n} \right )^{2n}}\le \frac{8}{\left(3 \right )^{2n}}$ which is a geometric series with ratio and so converges. The original series must also converge

T" - 1 22

d) $\frac{\ln(n)}{n^{3/2}}\le \frac{n^{1/3}}{n^{3/2}}\le \frac{1}{n^{7/6}}$ as $\ln(n)\le n^{1/3}$ for sufficient large n

The last series converges via p-test (p=7/6) and so the original series must also converge

$\blacksquare$

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