- #1

Joshk80k

- 17

- 0

## Homework Statement

Is

[tex]\sum \frac{1}{2n(2n+1)}[/tex]

convergent or divergent?

(Note that the summation is from 1 to infinity)

## Homework Equations

[tex]\int f(x) dx = L[/tex], (range is from 1 to infinity)

IF

L = [tex]\infty[/tex], divergent

L < [tex]\infty[/tex], convergent.

## The Attempt at a Solution

I tried a number of tests, and this is the only convergence test that I am not sure about. I attempted the ratio and root tests, but both were inconclusive.

As for my attempt,

[tex]\int \frac{1}{2n(2n+1)}dx[/tex], using the partial fractions method to integrate,

[tex]\int \frac{1}{2n(2n+1)}dx = \int \frac{A}{2n} + \frac{B}{2n+1} [/tex]

[tex]1 = 2nA + 2nB + A[/tex]

Matching up coefficients,

[tex]A^0: 1 = A[/tex]

and

[tex]A^1: 0 = 2A + 2B,[/tex]

[tex]B = -A = -1[/tex]

[tex]\int \frac{1}{2n(2n+1)}dx = \int \frac{1}{2n} + \frac{-1}{2n+1} [/tex]

Using substitution,

[tex]\int \frac{1}{2n} = \frac{1}{2}ln(2n)[/tex]

and

[tex]\int \frac{-1}{2n+1} = -\frac{1}{2}ln(2n+1)[/tex]

So, adding these two together, and using properties of the natural log,

[tex]\frac{1}{2}ln(2n) + -\frac{1}{2}ln(2n+1) = \frac{1}{2}ln(\frac{2n}{2n+1})[/tex]

Now, inserting the bounds 1 and t (Where t is infinity),

[tex]\frac{1}{2}ln(\frac{2t}{2t+1}) - \frac{1}{2}ln(\frac{2}{3})[/tex]

Here is where I am a little stumped. I want to say "Hey, t is infinity, and at large values of t, we can ignore the excess numbers (In this case, 2t + 1 is just a tiny bit different than 2t, so we ignore the 1 to make the cancellation) and as a result, I'll have a finite value, and for this reason, the answer is convergent.

BUT, I have always been a little shaky on convergence tests - am I doing this correctly or am I wrong to make that assumption about the cancellation?