Evaluating Summation of an Infinite Series

In summary, the given limit evaluates to zero as the sum in the numerator approaches infinity and the limit in the denominator approaches infinity. The summation of the series can be approximated using an integral, and the value of the limit depends on how the summation behaves. If the summation follows a power law with a constant exponent, the limit will approach zero. However, if the summation behaves differently, the limit could approach a different value.
  • #1
jisbon
476
30
Homework Statement
Evaluate ##\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]##
Relevant Equations
NIL
Evaluate ##\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]##

Hello. So I'm solving this question at the moment. I know I'm supposed to find out the summation of this before being able to solve the solution:

##(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+... +(\frac {n}{n})^{1.5}##However, I can't seem to find any AP or GP series in the equation above. There's no fixed difference between ##\frac {1}{n}^{1.5}## and ##\frac {2}{n}^{1.5}## and ... as far as I can tell? Any ideas on how to proceed here? Thank you.
 
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  • #2
Can you define the entire series? You cannot expect any answer if only two summands are given. This could be anything.
 
  • #3
fresh_42 said:
Can you define the entire series? You cannot expect any answer if only two summands are given. This could be anything.
Edited for clarity. Thank you.
 
  • #4
jisbon said:
Edited for clarity. Thank you.
Your parentheses don't match. It looks like ##a_n = n^{-\frac{5}{2}}\sum_{k=1}^\infty k^{\frac{3}{2}}##. But every ##a_n## is already infinitely large, so ##n## doesn't matter.
 
  • #5
fresh_42 said:
Your parentheses don't match. It looks like ##a_n = n^{-\frac{5}{2}}\sum_{k=1}^\infty k^{\frac{3}{2}}##. But every ##a_n## is already infinitely large, so ##n## doesn't matter.
Hi there, I think I might have edited it wrongly. What I meant was something like: ##(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...##

Thanks
 
  • #6
jisbon said:
Hi there, I think I might have edited it wrongly. What I meant was something like: ##(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...##

Thanks
This only changes the power of ##n## to ##\frac{3}{2}##, but it is still a sum of infinitely large numbers. Each sequence member has a constant value ##n##, i.e. a constant factor. The rest is infinitely large. The limit ##n \to \infty## is irrelevant.
 
  • #7
fresh_42 said:
This only changes the power of ##n## to ##\frac{3}{2}##, but it is still a sum of infinitely large numbers. Each sequence member has a constant value ##n##, i.e. a constant factor. The rest is infinitely large. The limit ##n \to \infty## is irrelevant.
Hi, thanks for your reply.
Firstly, my bad for not seeing the question properly, I edited it again as shown
Not sure what you meant by your reply above, but shouldn't I be using a formula for the summation for:
##\sum_{a=1}^{n} (\frac {a}{n})^{1.5}##
 
  • #8
jisbon said:
Hi, thanks for your reply.
Firstly, my bad for not seeing the question properly, I edited it again as shown
Not sure what you meant by your reply above, but shouldn't I be using a formula for the summation for:
##\sum_{a=1}^{n} (\frac {a}{n})^{1.5}##
Firstly we have ##\left( \frac{k}{n} \right)^{1.5} = \frac{k^{1.5}}{n^{1.5}}## and secondly a sum ##\frac{a}{N}+\frac{b}{N}+\frac{c}{N}+\ldots## where all terms have the same denominator ##N=n^{1.5}##. With the distributive law we have ##\frac{1}{n^{1.5}} \cdot (1^{1.5}+2^{1.5}+3^{1.5}+ \ldots)##. The second factor is infinitely large, regardless of the value of ##\frac{1}{n^{1.5}}##. But if every sequence member is infinitely large, the limit will be as well.
 
  • #9
fresh_42 said:
##\frac {1}{n^{1.5}}⋅(1^{1.5}+2^{1.5}+3^{1.5}+…n^{1.5})##
Ok, so from what I understand, the above equation will sum to 1 eventually? Thanks
 
  • #10
jisbon said:
Ok, so from what I understand, the above equation will sum to 1 eventually? Thanks
This is not what I wrote. The sum isn't finite, there are infinitely many terms which sum up to infinity. The factor ##\frac{1}{n^{1.5}}## doesn't change that.
 
  • #11
fresh_42 said:
This is not what I wrote. The sum isn't finite, there are infinitely many terms which sum up to infinity. The factor ##\frac{1}{n^{1.5}}## doesn't change that.
Okay I think I do understand now. Since ##\frac {1}{n^{1.5}}⋅(1^{1.5}+2^{1.5}+3^{1.5}+…n^{1.5})## =## \infty ##,
##\lim_{n \rightarrow +\infty} \frac {1} {n} [(\frac {1}{n})^{1.5} + (\frac {2}{n})^{1.5} +(\frac {3}{n})^{1.5}+ (\frac {4}{n})^{1.5}+...+(\frac {n}{n})^{1.5}]## =
##\lim_{n \rightarrow +\infty} \frac {1} {n} [\infty]?##
 
  • #12
Sorry, you have edited your question so often that I missed when you turned an infinite sum into a finite one. I thought you meant ##\sum_{k=1}^\infty k^{1.5}##.

If the sum is finite, things are different. Currently you have
$$
a_n = \frac{1}{n} \cdot \frac{1}{n^{1.5}} \cdot \left( 1^{1.5}+ 2^{1.5}+\ldots +n^{1.5} \right)= \frac{1}{n^{2.5}} \sum_{k=1}^n k^{1.5}
$$
I don't know a formula for ##\sum_{k=1}^n k^{1.5}##, but it is obviously less than ##n \cdot n^{1.5}=n^{2.5}##, so ##1## is an upper bound for your limit. There are probably approximations for the sum. If it behaves like ##n^{2.5-c}## for some ##c## independent of ##n##, then the limit ##n \to \infty## will be zero. So all depends on how ##\sum_{k=1}^n k^{1.5}## behaves.
 
  • #13
It's not hard to get an approximate formula for ##\sum_{k=1}^n k^{1.5}##. Just replace the sum with an integral. You should also be able to get a subleading term by approximating the error you make by doing that.
 
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  • #14
Dick said:
It's not hard to get an approximate formula for ##\sum_{k=1}^n k^{1.5}##. Just replace the sum with an integral. You should also be able to get a subleading term by approximating the error you make by doing that.
fresh_42 said:
Sorry, you have edited your question so often that I missed when you turned an infinite sum into a finite one. I thought you meant ##\sum_{k=1}^\infty k^{1.5}##.

If the sum is finite, things are different. Currently you have
$$
a_n = \frac{1}{n} \cdot \frac{1}{n^{1.5}} \cdot \left( 1^{1.5}+ 2^{1.5}+\ldots +n^{1.5} \right)= \frac{1}{n^{2.5}} \sum_{k=1}^n k^{1.5}
$$
I don't know a formula for ##\sum_{k=1}^n k^{1.5}##, but it is obviously less than ##n \cdot n^{1.5}=n^{2.5}##, so ##1## is an upper bound for your limit. There are probably approximations for the sum. If it behaves like ##n^{2.5-c}## for some ##c## independent of ##n##, then the limit ##n \to \infty## will be zero. So all depends on how ##\sum_{k=1}^n k^{1.5}## behaves.
Am I supposed to use the Integer Power Sum Formula here?
If so,
Is ##\sum_{k=1}^n k^{1.5}## simply = ##\frac {n^{2.5} + (n+1)^{2.5}}{5}##?
 
  • #15
Notice this can be seen as a Riemann sum. Just need to find which and then compute the integral.
 
  • #16
jisbon said:
Am I supposed to use the Integer Power Sum Formula here?
If so,
Is ##\sum_{k=1}^n k^{1.5}## simply = ##\frac {n^{2.5} + (n+1)^{2.5}}{5}##?

Yes, that's what you get approximating by integrals. And it's not an equality, it's only a pretty good approximation to the sum.
 
  • #17
Dick said:
Yes, that's what you get approximating by integrals. And it's not an equality, it's only a pretty good approximation to the sum.
So by evaluating it can I take the answer as the approximation? Or is there an accurate way to determine the value?
WWGD said:
Notice this can be seen as a Riemann sum. Just need to find which and then compute the integral.
I searched up on Riemann sum and apparently it is a certain kind of approximation of an integral by a finite sum. This means I will integrate the graph ##y= x^1.5## from x=1 to infinity?
 
  • #18
jisbon said:
So by evaluating it can I take the answer as the approximation? Or is there an accurate way to determine the value?

I searched up on Riemann sum and apparently it is a certain kind of approximation of an integral by a finite sum. This means I will integrate the graph ##y= x^1.5## from x=1 to infinity?

The Riemann sum is how you derive your power sum formula. If you draw your power sum as rectangles they are an upper sum for the curve ##y=x^{1.5}## and a lower sum for the curve ##y=(x+1)^{1.5}##. You integrate up to ##n## and average the two to get your formula. It's plenty accurate enough to use to figure the limit. Follow fresh_42's clues. If you want to be rigorous you can note that the true sum is between the area of the two curves.
 
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Related to Evaluating Summation of an Infinite Series

1. What is the definition of an infinite series?

An infinite series is an expression that consists of an infinite number of terms, which are added together. It can be written in the form of a1 + a2 + a3 + ... + an + ... where an is the n-th term of the series.

2. How do you evaluate an infinite series?

To evaluate an infinite series, you need to find the sum of all the terms in the series. This can be done by using various methods such as the geometric series test, the telescoping series test, or the integral test. In some cases, it may be possible to simplify the series or use known summation formulas to find the sum.

3. What is the difference between a convergent and a divergent series?

A convergent series is one in which the sum of all the terms is a finite number, while a divergent series is one in which the sum of the terms is infinite or does not exist. A series can only be considered convergent or divergent if its terms approach a limit as n goes to infinity.

4. What is the significance of the partial sum of an infinite series?

The partial sum of an infinite series is the sum of a finite number of terms in the series. It is used to approximate the sum of an infinite series and determine whether the series is convergent or divergent. As the number of terms in the partial sum increases, it becomes a better approximation of the actual sum of the series.

5. When can you use the comparison test to determine the convergence of an infinite series?

The comparison test can be used to determine the convergence of a series when the terms of the series are positive and can be compared to the terms of a known convergent or divergent series. If the terms of the given series are smaller than the terms of a convergent series, then the given series is also convergent. If the terms of the given series are larger than the terms of a divergent series, then the given series is also divergent.

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