Determine whether the series converges or diverges

  • Thread starter tnutty
  • Start date
  • #1
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Homework Statement



Determine whether the series converges or diverges.

[tex]\sum[/tex](1+5^n) / (1+6^n)

Homework Equations



limit comparison test, or just the comparison test.


The Attempt at a Solution



[attempt #1]

I have a couple of ways of trying to proves this.

1) (1+5^n) / (1+6^n) < (1 + 5^n) / 6^n =

(1/6)^n + (5/6)^n , which converges to 1/5 + 5 = 26/5

so since the series is bounded by 0 <= (1+5^n)/(1+6^n) <=26/5
it converges? This might not necessarily be true because a function might fluctuate between
those points until infinity, but I think that only cosine have this property. Thus this series converges.


[attempt # 2]

(1+5^n) / (1+6^n) =

1 / (1+6^n) + (5/6)^n

I know that (5/6)^n converges to 5, so

= 1/(1+6^n) + 5

now I apply the comparison test to 1/(1+6^n)

1/(1+6^n) < 1/(6^n), and this converges to 0.2

so ,

[0<= 1/(1+6^n) <= 0.2 ] + 5

so it converges since 1/(1+6^n) is bounded.


Either 1 right?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
619


Yes, use a comparison test. The first attempt is better. But if you are going to give values for the sum you should probably give a value for the lower limit on n. Is it 0 or 1 or 2 or what?
 

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