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### Test for Convergence

```
Date: 8/20/96 at 17:28:15
From: Anonymous
Subject: Test for Convergence

I cannot solve this problem:

Test the following series for convergence.

sum {from k=0 to infinity} [{log(k+1)-log k}/tan^(-1) (2/k)]

Tan^(-1) is tangent inverse. Because -1 < tan^(-1) < 1, I used the
comparison test, but I did not achieve the goal.

```

```
Date: 8/21/96 at 15:49:36
From: Doctor Anthony
Subject: Test for Convergence

This series appears to diverge since the term when k -> infinity is
not zero.

When k->inf  term is ln(1+1/k)/tan^(-1)(2/k)

As this tends to 0/0 we can use l'Hopital's rule to find its limiting
value.

Differentiating top and bottom separately we get:

{1/(1+1/k)}(-1/k^2)
-------------------
{1/(1+4/k^2)}(-2/k^2)

= (1/2){k/(k+1)}
--------------
k^2/(k^2+4)

=  (1/2)k(k^2+4)
-------------
k^2(k+1)

= (1/2)(k^2+4)
------------
k^2 + k

= (1/2)(1+4/k^2)
--------------
1 + 1/k

If we let k->inf. This ratio -> 1/2

We can only get convergence if the terms tend to zero as k-> infinity,
and this is not happening, so the series diverges.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Sequences, Series

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