MHB Which Tests Determine Convergence for These Series?

Click For Summary
The discussion focuses on determining the convergence of two series involving factorials and exponential functions. Users are encouraged to apply convergence tests to the given series rather than repeatedly asking similar questions. There is a note about the difficulty of interpreting the series due to the lack of LaTeX formatting, which affects clarity. A moderator renumbers the problems for better organization, clarifying that what was referred to as #3 is actually #6. The conversation emphasizes the importance of clear notation in mathematical discussions.
Tebow15
Messages
10
Reaction score
0
Test these for convergence.

5.
infinity
E...((n!)^2((2n)!)^2)/((n^2 + 2n)!(n + 1)!)
n = 0

6.
infinity
E...(1 - e ^ -((n^2 + 3n))/n)/(n^2)
n = 3

note: for #3: -((n^2 + 3n))/n) is all to the power of e

Btw, E means sum.

Which tests should I use to solve these?
 
Last edited by a moderator:
Physics news on Phys.org
Rather than posing a series of similar questions, you should apply the help given to you on one thread to attempt the other problems.
 
(Poolparty)
 
cacophony said:
Test these for convergence.

5.
infinity
E...((n!)^2((2n)!)^2)/((n^2 + 2n)!(n + 1)!)
n = 0

6.
infinity
E...(1 - e ^ -((n^2 + 3n))/n)/(n^2)
n = 3

note: for #3: -((n^2 + 3n))/n) is all to the power of e

What do you mean by #3 ? , maybe you meant the question you posted earlier. Not writing the sums in LaTeX makes it so difficult to interpret !
 
ZaidAlyafey said:
What do you mean by #3 ? , maybe you meant the question you posted earlier. Not writing the sums in LaTeX makes it so difficult to interpret !

Moderator note: I renumbered the problems so as to have no more than two problems in a thread. The #3 should be a #6. It's Problem #6.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 17 ·
Replies
17
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
3
Views
1K