Answer Limit of a Series: Sum of 1+(2)^n/(3^n) from 1 to ∞

  • Thread starter Thread starter fk378
  • Start date Start date
  • Tags Tags
    Limit Series
fk378
Messages
366
Reaction score
0

Homework Statement


Determine whether the series is convergent or divergent. If convergent, find its sum.

sum of [1+(2)^n] / (3^n)
from 1 to inf


Homework Equations


I know that the sum of a geometric series is 1/(1-r)


The Attempt at a Solution


The sum of a series is the limit of its partial sums.
I separate the summation into 2 parts: 1/(3^n) + (2^n)/(3^n)

I can see from this that the limits of both of these approach 0, so I conclude that the sum the series is 0.

However, my book says the answer is 5/2 and I tried to solve this a different way and got 5/2 as well. I re-wrote the separate summations as (1/3)^n + (2/3)^n and notice the ratio, r, is 1/3 and 2/3, respectively. Applying the "relevant equation" of 1/(1-r) I solve the summations and get 5/2.

However, if the sum of a series is the limit of its partial sums, why am I getting a different value for my first attempt?
 
Physics news on Phys.org
You are confusing the limit of a sequence with the sum of a series. That's all.
 
My book says that "the sum of a series is the limit of the sequence of partial sums."
 
You didn't do any partial sums. You just looked at the limits of the individual terms.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top