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

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SUMMARY

The series defined as An=Ʃ(k)/[(n^2)+k], where the summation runs from k=0 to n, converges but does not have a simple closed-form expression for its sum. The convergence test applied, mod(An+1/An), yielded a value of 1, indicating convergence. However, the actual evaluation of the series requires using the Squeeze Theorem to bracket the sum between two limits. The result of the summation can be expressed in terms of the Psi function, specifically as n+1+n^2 Ψ(n^2)-n^2 Ψ(n^2+n+1).

PREREQUISITES
  • Understanding of series convergence tests, specifically the ratio test.
  • Familiarity with the Squeeze Theorem in calculus.
  • Knowledge of the Psi function (di-Gamma function) and its properties.
  • Basic proficiency in mathematical notation and summation techniques.
NEXT STEPS
  • Study the properties and applications of the Psi function in mathematical analysis.
  • Learn about advanced convergence tests for series beyond the ratio test.
  • Explore the Squeeze Theorem with practical examples to solidify understanding.
  • Investigate comparison series and their role in evaluating convergence of complex series.
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Mathematicians, students studying advanced calculus, and anyone interested in series convergence and evaluation techniques.

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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