Proving the convergence of series

In summary: Thanks!In summary, the series converges if the absolute value of r is smaller than 1, diverges if the absolute value of r is greater or equal to 1, and is known to converge if p is greater than 1.
  • #1
Raymondyhq
8
0

Homework Statement


Prove the convergence of this series using the Comparison Test/Limiting Comparison Test with the geometric series or p-series. The series is:

The sum of [(n+1)(3^n) / (2^(2n))] from n=1 to positive ∞

The question is also attached as a .png file

2. Homework Equations

The geometric series with a * r^n is known to:
  • converge if the absolute value of r is smaller than 1
  • diverge if the absolute value of r is greater or equal to 1
The p-series (1/n^p) is known to:
  • converge if p is greater than 1
  • divergent if otherwise
Please refer to this website for the definition of the Comparison Test and the Limiting Comparison Test:

The Attempt at a Solution


This is as far as I got:
∑ [(n+1)(3^n) / 2^(2n)] can be split into two series,
∑ [n(3^n) / (4^n)] + ∑ [(3^n) / (4^n)]
The latter series is known to converge because it is a geometric series with r = 3/4
However, I am stuck trying to solve the first series.

Using the Ratio Test to determine whether the series converges or diverges is quite simple, and I have worked it out.

By the way, could anyone tell me how to make the formulas look more ... natural? Instead of using ^ and / as well as a ton of brackets.

Any help is appreciated!
 

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  • #2
Can you trap ##(n+1)(\dfrac{3}{4})^n## and a geometric series ##\sum q^n\,##?
 
  • #3
fresh_42 said:
Can you trap ##(n+1)(\dfrac{3}{4})^n## and a geometric series ##\sum q^n\,##?
Could you elaborate? I don't quite understand what you mean by ∑qn
 
  • #4
What is a geometric series? You're requested to use one. So you need to know what it is, which of them converge and finally what comparison means. If you can answer all three questions correctly, you're almost there.
 
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  • #5
Raymondyhq said:
∑ [n(3^n) / (4^n)] + ∑ [(3^n) / (4^n)]
The first series can be written as ##\sum n (3/4)^n##. What can you say about this series by using the Comparison Test or Limit Comparison Test?
 
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  • #6
fresh_42 said:
What is a geometric series? You're requested to use one. So you need to know what it is, which of them converge and finally what comparison means. If you can answer all three questions correctly, you're almost there.
Mark44 said:
The first series can be written as ##\sum n (3/4)^n##. What can you say about this series by using the Comparison Test or Limit Comparison Test?
This is exactly what I am having trouble solving. I could try to move n into the brackets with 3/4.
∑n(3/4)n = ∑(n√n)n(3/4)n = ∑(n√n 3/4)n
Then I am stuck.
 
  • #7
You have to find a quotient ##q## with ##(n+1)\left( \dfrac{3}{4} \right)^n < q^n##. Any ideas how to find a ##q\,##? Hint: find a quotient ##p## with ##(n+1) < p^n## first.
 
  • #8
fresh_42 said:
You have to find a quotient ##q## with ##(n+1)\left( \dfrac{3}{4} \right)^n < q^n##. Any ideas how to find a ##q\,##? Hint: find a quotient ##p## with ##(n+1) < p^n## first.
I am having trouble finding a p that satisfies pn>(n+1)
:cry:
 
  • #9
Raymondyhq said:
I am having trouble finding a p that satisfies pn>(n+1)
:cry:
In this case the quotient would need to be ... greater than 1?
 
  • #10
Raymondyhq said:
I am having trouble finding a p that satisfies pn>(n+1)
Well, ##p=2## would certainly do, for all ##n>2##.
Raymondyhq said:
In this case the quotient would need to be ... greater than 1?
Exactly. Now in the example with ##p=2## we will lose too much space, because we need ##p^n \cdot \left( \dfrac{3}{4} \right)^n < q^n < 1##.
This narrows it down, but there are still enough possibilities.
 
  • #11
What about p = 11/10 ? pn would be greater than (n+1) if n is sufficiently large, and pn * (3/4)n = (33/40)n which would satisfy r < 1.
 
  • #12
Raymondyhq said:
What about p = 11/10 ? pn would be greater than (n+1) if n is sufficiently large, and pn * (3/4)n = (33/40)n which would satisfy r < 1.
Yes, that would do. You still need to prove, that it gets larger than ##n+1## (binomial formula) for which (?) ##n## and sweep everything together for a formal proof.
 
  • #13
fresh_42 said:
Yes, that would do. You still need to prove, that it gets larger than ##n+1## (binomial formula) for which (?) ##n## and sweep everything together for a formal proof.

I could set up the equation (11/10)n = n+1 but I would not be able to solve it without the aid of a graphing calculator. Using a graphing calculator, I found that for all x > 40, pn is greater than n+1. Thus, I would have proved that when n goes from 1 to positive infinity, the series (33/40)n > (n+1)(3/4)n.

Could you suggest a simpler method? In my Calculus 2 course, the use of the binomial formula was not mentioned.

Thanks!
 
  • #14
Well, you have to explain why ##n+1 < \left(\dfrac{11}{10} \right)^n ## for large ##n\;[*]##. I find it easiest by using ##\left(\dfrac{11}{10} \right)^n = \left(1+\left( \dfrac{1}{10}\right) \right)^n=1+\binom{n}{1}\left(\dfrac{1}{10} \right)+\binom{n}{2}\left(\dfrac{1}{10} \right)^2+ \ldots + \left(\dfrac{1}{10} \right)^{11}## but you can as well proof that ##x \longmapsto (x+1)^{\frac{1}{x}}## is monotone decreasing with limit ##1##, or by induction that ##(n+1)^{\frac{1}{n}} > (n+2)^{\frac{1}{n+1}}## or rule out by another method, that it cannot increase for large ##n## at some point. Maybe it's even enough to simply say, that exponential behavior outnumbers polynomial behavior, or just take it for granted - I don't know. For a rigorous proof, we need ##(n+1) \cdot \left( \dfrac{3}{4} \right)^n {<}_{[*]} \left(\dfrac{11}{10} \right)^n \cdot \left( \dfrac{3}{4} \right)^n = \left( \dfrac{33}{40} \right)^n## and the value of ##\sum_{n \in \mathbb{N}} \left( \dfrac{33}{40} \right)^n##. Thus some argument for ##[*]## is needed, in my opinion.
 
Last edited:
  • #15
Raymondyhq said:

Homework Statement


Prove the convergence of this series using the Comparison Test/Limiting Comparison Test with the geometric series or p-series. The series is:

The sum of [(n+1)(3^n) / (2^(2n))] from n=1 to positive ∞

The question is also attached as a .png file

2. Homework Equations

The geometric series with a * r^n is known to:
  • converge if the absolute value of r is smaller than 1
  • diverge if the absolute value of r is greater or equal to 1
The p-series (1/n^p) is known to:
  • converge if p is greater than 1
  • divergent if otherwise
Please refer to this website for the definition of the Comparison Test and the Limiting Comparison Test:

The Attempt at a Solution


This is as far as I got:
∑ [(n+1)(3^n) / 2^(2n)] can be split into two series,
∑ [n(3^n) / (4^n)] + ∑ [(3^n) / (4^n)]
The latter series is known to converge because it is a geometric series with r = 3/4
However, I am stuck trying to solve the first series.

Using the Ratio Test to determine whether the series converges or diverges is quite simple, and I have worked it out.

By the way, could anyone tell me how to make the formulas look more ... natural? Instead of using ^ and / as well as a ton of brackets.

Any help is appreciated!
Raymondyhq said:
I could set up the equation (11/10)n = n+1 but I would not be able to solve it without the aid of a graphing calculator. Using a graphing calculator, I found that for all x > 40, pn is greater than n+1. Thus, I would have proved that when n goes from 1 to positive infinity, the series (33/40)n > (n+1)(3/4)n.

Could you suggest a simpler method? In my Calculus 2 course, the use of the binomial formula was not mentioned.

Thanks!

Have you had the fact that
$$\lim_{x \to \infty} \frac{\ln (x)}{x} = 0 ?$$
If so, that means that for any fixed positive constant ##c = \ln(1+\alpha) ## (with ##\alpha > 0##) there is some ##N_{\alpha}## such that ##\ln(x) < c x## for ##x > N_{\alpha}##, hence ##n < e^{nc} = (1+\alpha)^n ## for ##n > N_{\alpha}##. Therefore, for ##0 < r < 1## we have ##n r^n < (r(1+\alpha))^n## for ##n > N_{\alpha}##. As long as you choose ##\alpha## so that ##(1+\alpha) r \equiv \rho < 1## your terms ##n r^n## are bounded above by a convergent geometric sequence ##\rho^n## for all ##n## beyond a certain point. For given ##r## and ##\rho \in (r,1)## you can figure out what values of ##\alpha## will work, then pick one such value and thus figure out what is ##N_{\alpha}##. However, for just proving the result you do not need to do any numerics; it is enough to know that certain things exist (and, in principle, could be found explicitly if need be).
 

Related to Proving the convergence of series

What is the definition of convergence of a series?

The convergence of a series is a mathematical concept that describes the behavior of an infinite sequence of numbers as the number of terms in the sequence increases. A series is said to be convergent if the terms in the sequence approach a finite limit as the number of terms increases.

How do you prove the convergence of a series?

There are various methods for proving the convergence of a series, including the comparison test, the ratio test, and the root test. These tests involve examining the behavior of the terms in the series and determining if they approach a finite limit or not.

What is the difference between absolute convergence and conditional convergence?

A series is said to be absolutely convergent if the sum of the absolute values of its terms is finite. On the other hand, a series is said to be conditionally convergent if it is convergent, but not absolutely convergent. In other words, in a conditionally convergent series, the sum of the absolute values of the terms would be infinite, but the series still converges to a finite limit.

Can a series be convergent and divergent at the same time?

No, a series cannot be both convergent and divergent at the same time. If a series converges, it means that the terms in the sequence approach a finite limit, while if a series diverges, it means that the terms in the sequence do not approach a finite limit. These two concepts are mutually exclusive.

What is the importance of proving the convergence of a series?

Proving the convergence of a series is important because it allows us to determine the behavior of infinite sequences of numbers and to calculate their sums. This is crucial in various fields of mathematics, physics, and engineering, where infinite series are used to represent real-world phenomena and make predictions.

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