# Finding convergence of this series using Integral/Comparison

• Rijad Hadzic
In summary, convergence refers to the behavior of a sequence or series as its terms continue to be added and can either converge or diverge. The integral and comparison tests are methods for determining the convergence or divergence of a series, with the integral test using a related improper integral and the comparison test comparing the series to a known series. However, these tests can only be used for certain series with non-negative terms and have limitations in providing an exact value for the sum of the series and may be inconclusive for some series.
Rijad Hadzic

## Homework Statement

series from n = 1 to infinity, (ne^(-n))

## The Attempt at a Solution

I want to use integral test.
I know this function is:
positive (on interval 1 to infinity)
continous
and finding derivative of f(x) = xe^(-x) I found it to be ultimately decreasing.

So integral test is applicable.

I set up integral from 1 to infinity (xe^(-x))

u = x du = dx
v = -e^(-x) dv = e^(-x)

-xe^(-x) + integral e^(-x)

-xe^(-x) - e^(-x) = -e^(-x) (x + 1)

evaluating from 1 to t

$-(1/e^t) (t+1) + 2/e^(1)$

but now when I do lim t -> infinity, -(1/e^t) (t+1) should = infinity/infinity, which would mean $a_n$ would be divergent, but it is convergent.

does anyone know where my mistake is??

Rijad Hadzic said:

## Homework Statement

series from n = 1 to infinity, (ne^(-n))

## The Attempt at a Solution

I want to use integral test.
I know this function is:
positive (on interval 1 to infinity)
continous
and finding derivative of f(x) = xe^(-x) I found it to be ultimately decreasing.

So integral test is applicable.

I set up integral from 1 to infinity (xe^(-x))

u = x du = dx
v = -e^(-x) dv = e^(-x)-xe^(-x) + integral e^(-x)

-xe^(-x) - e^(-x) = -e^(-x) (x + 1)

evaluating from 1 to t

$-(1/e^t) (t+1) + 2/e^(1)$
I get something slightly different from what you got, namely ##te^{-t} - e^{-t} + 2e^{-1} = (t - 1)e^{-t} + 2e^{-1}##
Rijad Hadzic said:
but now when I do lim t -> infinity, -(1/e^t) (t+1) should = infinity/infinity
But the fact that you're getting the indeterminate form ##[\frac \infty \infty]## doesn't tell you anything. You can use L'Hopital's Rule to actually evaluate your limit, which results in an actual value for this limit.
Rijad Hadzic said:
, which would mean $a_n$ would be divergent, but it is convergent.

does anyone know where my mistake is??

BTW, this thread was marked as "Solved" but that didn't appear to really be the case, so I have changed it to "Unsolved."

## 1. What is convergence?

Convergence is a mathematical concept that refers to the behavior of a sequence or series as its terms continue to be added. If the terms of the sequence or series approach a finite limit as more terms are added, it is said to converge. If the terms do not approach a finite limit, it is said to diverge.

## 2. How can we test for convergence of a series using integration?

The integral test is a method for determining the convergence or divergence of an infinite series by comparing it to a related improper integral. If the integral converges, then the series also converges. If the integral diverges, then the series also diverges.

## 3. What is the comparison test for convergence of a series?

The comparison test is a method for determining the convergence or divergence of a series by comparing it to a known series with known convergence or divergence. If the known series converges and the given series is always less than or equal to the known series, then the given series also converges. If the known series diverges and the given series is always greater than or equal to the known series, then the given series also diverges.

## 4. Can the integral and comparison tests be used for all series?

No, the integral and comparison tests can only be used for series with non-negative terms. Also, the series must be continuous and decreasing for the integral test, and the series must be positive and comparable to another series for the comparison test.

## 5. What are the limitations of using the integral and comparison tests for convergence?

The integral and comparison tests can only determine whether a series converges or diverges, but they cannot provide an exact value for the sum of the series. Additionally, these tests may be inconclusive for some series, and other convergence tests may be needed to determine the convergence or divergence of the series.

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