Series Convergence: An=Ʃ(k)/[(n^2)+k] - Find Value

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Homework Help Overview

The discussion centers around the convergence of the series defined by An=[Ʃ(k)/[(n^2)+k]], where the sum is taken from k=0 to n. Participants are exploring the value to which this series converges.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Some participants attempt convergence tests, noting that the value obtained is 1. Others suggest using the squeeze theorem to evaluate the limit of the sum. There is a distinction made between finding the limit of the function and the sum of the series itself.

Discussion Status

Participants are actively discussing the nature of the series and the challenges in evaluating it. Some guidance has been offered regarding the use of comparison series and the limitations of elementary functions in finding a closed form for the sum.

Contextual Notes

There is an emphasis on the fact that the upper limit n is part of the expression for the terms being summed, which complicates the evaluation. The discussion also highlights that the summation cannot be performed in elementary terms.

oneomega
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Homework Statement



An=Ʃ(k)/[(n^2)+k]
the sum is k=0 to n, the question is, to which value does the this series converge to

Homework Equations


i know for sure that this series converges, but could not figure out the value to whch it converges


The Attempt at a Solution



i did the convergence test, mod(An+1/An).. the value is 1. what i do now?
 
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oneomega said:

Homework Statement



An=Ʃ(k)/[(n^2)+k]
the sum is k=0 to n, the question is, to which value does the this series converge to

Homework Equations


i know for sure that this series converges, but could not figure out the value to whch it converges

The Attempt at a Solution



i did the convergence test, mod(An+1/An).. the value is 1. what i do now?


You want to use the squeeze theorem. Try and bracket that sum between two sums whose limit you can evaluate and whose limit turn out to be the same. Here's a hint. The k in the denominator is what's making it hard to evaluate.
 
actaully i don't want to find the limit of the fn. i want the sum of the series.
 
oneomega said:
actaully i don't want to find the limit of the fn. i want the sum of the series.

'converges' means you take the limit of the sums ##A_n## as n->infinity. There's no simple closed form expression for the sum. There is a simple expression for the limit of the sum. This isn't really the same as the usual infinite sum problem. ##A_n## isn't the partial sum of some series. The upper limit n is in the expression for the terms you are summing.
 
Last edited:
oneomega said:

Homework Statement



An=Ʃ(k)/[(n^2)+k]
the sum is k=0 to n, the question is, to which value does the this series converge to

Homework Equations


i know for sure that this series converges, but could not figure out the value to whch it converges


The Attempt at a Solution



i did the convergence test, mod(An+1/An).. the value is 1. what i do now?


Using the "test"
\left|A_n + \frac{1}{A_n} \right|
(which is what you WROTE) will get you nowhere. Even the correct test ##A_{n+1}/A_n## (written as A_{n+1}/A_n or A_(n+1)/A_n) will still get you nowhere: your problem is NOT to decide on convergence of an infinite series.

The summation cannot be performed in elementary terms: Maple gets the answer
\sum_{k=0}^n \frac{k}{n^2+k} = n+1+n^2 \Psi(n^2)-n^2 \Psi(n^2+n+1)
where ##\Psi(x)## is the so-called "Psi function" or "di-Gamma function", defined as the logarithmic derivative of the Gamma function ##\Gamma(x)##:
\Psi(x) \equiv \frac{d \ln(\Gamma(x))}{dx} = \frac{\Gamma\, ^{\prime} (x)}{\Gamma(x)}.
 
Last edited:
@ray wickson,
could you tell me an answer for this.
 
oneomega said:
@ray wickson,
could you tell me an answer for this.

Nobody is going to 'tell you an answer', you have to work on it. You really don't need any nonelementary functions for this. Just think about comparison series.
 

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