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Divergent sum

  • Thread starter jisbon
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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}]##
Homework 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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fresh_42

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Can you define the entire series? You cannot expect any answer if only two summands are given. This could be anything.
 
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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.
 

fresh_42

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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.
 
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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
 

fresh_42

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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.
 
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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}##
 

fresh_42

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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.
 
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##\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
 

fresh_42

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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.
 
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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]?##
 

fresh_42

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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.
 

Dick

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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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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.
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}##?
 

WWGD

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Notice this can be seen as a Riemann sum. Just need to find which and then compute the integral.
 

Dick

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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.
 
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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?
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?
 

Dick

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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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