Question

The sum of n root 2 from 1 to infinity.
I need desperately help with series! Please tell me you know how to find whether this is convergent or divergent?

Answer 1

\[\sqrt[n]{2}\]

Answer 2

the limit does not exist or it is infinite, then we say that the improper integral is divergent.

Answer 3

Why does the limit not exist? As n approaches infinity, doesn't the value get closer to zero?

Answer 4

Yes, but it will never get to zero. That's what makes it infinite

Answer 5

But a limit doesn't ever actually REACH something, does it? It just gets reallllllly close.

Answer 6

Exactly. Even though you are GETTING closer, you will NEVER reach zero

Answer 7

Okay, wait. Let me give you another question then.
\[\sum_{n=1}^{\infty} \frac{ 1 + 2^{n} }{3^{n}}\]

Answer 8

How do I find if this is divergent or convergent?
L'hospital's doesn't work, dividing by the highest power doesn't work, can't take the limit. So what else do I do?

Answer 9

http://www.sosmath.com/calculus/improper/convdiv/convdiv.html

Answer 10

I'm sorry I'm struggling so much with this, but I just don't understand. I know one way to test for divergence is the divergence test. Meaning, if the limit of the series DNE or does not equal 0, then the series is divergent.
However, a series CAN add up to a number other than 0 and still be convergent. What am I missing here?

Answer 11

If anything is divergent, it just means it will never touch zero
The sum of n root 2 from 1 to "infinity"
In your statement infinity tells it all

Answer 12

But all of my series go to infinity and some of them ARE convergent.

Answer 13

For example: \[\sum_{n=1}^{\infty} \frac{ 1 }{ \sqrt{2}^{n}}\] is convergent.

Answer 14

In the first example, perhaps it would be useful to compare it to 1^(1/n).
In your second example,
\( \displaystyle \sum_{n=1}^{\infty} \dfrac{1 + 2^n}{3^n} \)
If you understand that 1^n = 1, you might notice the opportunity for two geometric series here.

Answer 15

Where \( \sqrt[n]{2} = 2^{1/n} \) by definition. (If that doesn't make sense, think \( \left(2^{1/n}\right)^n = \left(\sqrt[n]{2}\right)^{n} \)... they both cancel the operation, right?)