Answer to Question #281792 in Calculus for chaitu

Question #281792

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

.


1
Expert's answer
2021-12-29T12:58:35-0500

24.

n=1(n!)2n2n∑︁^∞_{n=1}\frac{(n!)^2}{n^{2n}}


limnn!nn=limn123...(n1)nnn...n=0\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,

limnunn=limn(n!)2n2nn=0<1\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=1n3+12n+1∑︁^∞_{n=1}\frac{n^3+1}{2^n+1}


using L'Hopital's rule:

limnn3+12n+1=limn62nln32=0\displaystyle \lim_{n\to \infin}\frac{n^3+1}{2^{n}+1}=\displaystyle \lim_{n\to \infin}\frac{6}{2^nln^32}=0


then:

limnunn=limnn3+12n+1n=0<1\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.

1234+2345+3456+4567+...=n=1n(n+1)(n+2)(n+3)\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 n(n+1)(n+2)(n+3)<1n2\frac{n}{(n+1)(n+2)(n+3)}<\frac{1}{n^2} and series 1n2\sum \frac{1}{n^2} converges, then series


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


27.

21331+31431+41531+...=n=1n1(n+1)31\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}


un=n1(n+1)31,vn=1n5/2u_n=\frac{\sqrt n-1}{(n+1)^3-1},v_n=\frac{1}{n^{5/2}}


limn(un/vn)=10\displaystyle \lim_{n\to \infin}(u_n/v_n)=1\neq 0


we know that series n=11n5/2\sum^{\infin}_{n=1}\frac{1}{n^{5/2}} converges, so series n=1n1(n+1)31\sum^{\infin}_{n=1}\frac{\sqrt n-1}{(n+1)^3-1} converges as well


28.

21p+32p+43p+...=n=1n+1np\frac{2}{1^p}+\frac{3}{2^p}+\frac{4}{3^p}+...=\sum^{\infin}_{n=1}\frac{n+1}{n^p}


un=n+1np,vn=1np1u_n=\frac{n+1}{n^p},v_n=\frac{1}{n^{p-1}}


limn(un/vn)=10\displaystyle \lim_{n\to \infin}(u_n/v_n)=1\neq 0


then:

since series n=11np1\sum^{\infin}_{n=1}\frac{1}{n^{p-1}} diverges if p ≤ 2 and converges if p > 2,


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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment

LATEST TUTORIALS
New on Blog