Question

Test the convergence of ∑︁∞

n=1

[︂

n!2n

2

n

]︂

.

Ans: ∑︁un Converges

25. Test the convergence of ∑︁∞

n=1

[︂

n

3 + 1

2

n + 1]︂

.

Ans: ∑︁un Converges

Hint: Use vn as

1

√n

.

26. Test the convergence of

1

2 · 3 · 4

+

2

3 · 4 · 5

+

3

4 · 5 · 6

+

4

5 · 6 · 7

+ · · · .

Ans: ∑︁un Converges

Hint: Use vn as

1

n2

.

27. Test the convergence of

√

2 − 1

3

3 − 1

+

√

3 − 1

4

3 − 1

+

√

4 − 1

5

3 − 1

+ · · · .

Ans: ∑︁un Converges

Hint: Use vn as

1

n

5

2

.

28. Test the convergence of

2

1

p +

3

2

p +

4

3

p + · · · .

Ans: ∑︁un Converges if p > 2 and ∑︁un Diverges if p ≤ 2

Hint: Use vn as

1

n

p−1

.

Answer 1

24.

$∑︁^∞_{n=1}\frac{(n!)^2}{n^{2n}}$

$\displaystyle \lim_{n\to \infin}\frac{{n!}}{n^n}=\displaystyle \lim_{n\to \infin}\frac{{1\cdot2\cdot3...(n-1)n}}{n\cdot n...n}=0$

so,

$\displaystyle \lim_{n\to \infin}\sqrt[n]{u_n}=\displaystyle \lim_{n\to \infin}\sqrt[n]{\frac{(n!)^2}{n^{2n}}}=0<1$

so, series converges

25.

$∑︁^∞_{n=1}\frac{n^3+1}{2^n+1}$

using L'Hopital's rule:

$\displaystyle \lim_{n\to \infin}\frac{n^3+1}{2^{n}+1}=\displaystyle \lim_{n\to \infin}\frac{6}{2^nln^32}=0$

then:

$\displaystyle \lim_{n\to \infin}\sqrt[n]{u_n}=\displaystyle \lim_{n\to \infin}\sqrt[n]{\frac{n^3+1}{2^{n}+1}}=0<1$

so, series converges

26.

$\frac{1}{2 · 3 · 4}+\frac{2}{3 · 4 · 5}+\frac{3}{4 · 5 · 6}+\frac{4}{5 · 6 · 7}+...=\sum_{n=1}^{\infin}\frac{n}{(n+1)(n+2)(n+3)}$

since $\frac{n}{(n+1)(n+2)(n+3)}<\frac{1}{n^2}$ and series $\sum \frac{1}{n^2}$ converges, then series

$\sum_{n=1}^{\infin}\frac{n}{(n+1)(n+2)(n+3)}$ converges as well

27.

$\frac{\sqrt 2-1}{3^3-1}+\frac{\sqrt 3-1}{4^3-1}+\frac{\sqrt 4-1}{5^3-1}+...=\sum^{\infin}_{n=1}\frac{\sqrt n-1}{(n+1)^3-1}$

$u_n=\frac{\sqrt n-1}{(n+1)^3-1},v_n=\frac{1}{n^{5/2}}$

$\displaystyle \lim_{n\to \infin}(u_n/v_n)=1\neq 0$

we know that series $\sum^{\infin}_{n=1}\frac{1}{n^{5/2}}$ converges, so series $\sum^{\infin}_{n=1}\frac{\sqrt n-1}{(n+1)^3-1}$ converges as well

28.

$\frac{2}{1^p}+\frac{3}{2^p}+\frac{4}{3^p}+...=\sum^{\infin}_{n=1}\frac{n+1}{n^p}$

$u_n=\frac{n+1}{n^p},v_n=\frac{1}{n^{p-1}}$

$\displaystyle \lim_{n\to \infin}(u_n/v_n)=1\neq 0$

then:

since series $\sum^{\infin}_{n=1}\frac{1}{n^{p-1}}$ diverges if p ≤ 2 and converges if p > 2,

same thing for $\sum^{\infin}_{n=1}\frac{n+1}{n^p}$ : diverges if p ≤ 2 and converges if p > 2