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Convergent or divergent?

  • Thread starter tnutty
  • Start date
327
1
1. Homework Statement

Determine whether the series converges or diverges.

[tex]\sum[/tex] 3+7n / 6n




Attempt :

Comparison test :

3+7n / 6n < 7n / 6n

3+7n / 6n < (6/7)n

since (6/7)n is a geometric series and is convergent is
3+7n / 6n convergent as well?
 
327
1
I see my mistake

3+7n / 6n > (7/6)^n

but then what?
 
316
0
Hint: For a series
[tex]\sum_{n=0}^{\infty}a_n[/tex]

to converge, the terms have to converge to zero, i.e.

[tex]\lim_{n->\infty}a_n=0[/tex].
 
327
1
I could use the limit comparison test but I get stuck.
 
316
0
You have [tex]a_n=3+7^n/6^n>3[/tex], so can [tex]a_n[/tex] converge to zero?
 
327
1
How did you figure that inequality ?

(7/6)^n converges to -7
 
316
0
327
1
(7/6)^n

This is a geometric series. And I know that by definition a*r^(n-1) ,where |r|<1 = a/(1-r).

so (7/6)^n
=

(7/6) * (7/6)^(n-1)

so,
a = 7/6
r = 7/6

it follows that

(7/6)^n = a/(1-r) = (7/6) / (1-7/6) = -7
 
327
1
wait I see what your saying. How comes this the above statement is wrong?
 
327
1
no your right, 7/6 > 1 so this series diverges.
Thanks
 

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