Series Homework Question: Divergent or Convergent? Methods Compared

[Of Mike and Men]

Homework Statement

I was watching a PatrickJMT video on ratio test and he gave this problem. I solved it before he did, and he got that it was divergent. He didn't simplify it initially, so our methods of approach are different. Did I do something wrong? I checked my calculator to make sure all my simplifications were correct. So I'm assuming I did something near the limit portion that was incorrect (if at all). Thanks

∞
∑ n/3¹⁺³ⁿ
n=1

Homework Equations

The Attempt at a Solution

∞
∑ n/3¹⁺³ⁿ = n/(3⋅27ⁿ)
n=1

lim |(n+1)/(3⋅27ⁿ⁺¹) / (n/3 *27ⁿ)|
n→∞

= lim |(n+1)/n| * (1/27)
n→∞
= 1 * 1/27
=1/27 < 1

Convergent by Ratio Test

 

[member 587159]

Did you use the ratio test correctly?

 

[Of Mike and Men]

Did you use the ratio test correctly?

Nope. Got it, thanks.

[edit]

Actually I typed my work incorrectly. I'll fix it. but to be sure...

 

[Ray Vickson]

Homework Statement

I was watching a PatrickJMT video on ratio test and he gave this problem. I solved it before he did, and he got that it was divergent. He didn't simplify it initially, so our methods of approach are different. Did I do something wrong? I checked my calculator to make sure all my simplifications were correct. So I'm assuming I did something near the limit portion that was incorrect (if at all). Thanks

∞
∑ n/3¹⁺³ⁿ
n=1

Homework Equations

The Attempt at a Solution

∞
∑ n/3¹⁺³ⁿ = n/(3⋅27ⁿ)
n=1

lim |(n+1)/(3⋅27ⁿ⁺¹) / (n/3 *27ⁿ)|
n→∞

= lim |(n+1)/n| * (1/27)
n→∞
= 1 * 1/27
=1/27 < 1

Convergent by Ratio Test

It is convergent, as you have shown correctly using the ratio test.

You can actually find the sum $\sum_{n=1}^N n/3^{3n+1}$ explicitly (as a reasonably simple closed-from expression in $N$), then take the limit as $N \to \infty$. All you need to do is find a formula for $f_N(x) =\sum_{n=1}^N n x^n$, and these are widely available, or you can work it out as $f_N(x) = x (d/dx) g_N(x)$, where $g_N(x) = \sum_{n=1}^N x^n$, which is elementary.