How to Find the Partial Sum of a Series?

[Feynmanfan]

Hello everybody!

I'm having some trouble with series. My calculus teacher asked us to find the partial sum of

Sigma from 1 to n [n^-(1 + 1/n)]

It is obvious that the series diverges when trying to find the infinite sum. However, is it possible to find an expression dependent of n of the partial sum? I don't know where to start from

 

[HallsofIvy]

Well, one problem you have is that your problem doesn't make sense.

You shouldn't use "n" for both the upper limit of summation and as the index inside the sum.

I assume that what you really mean is
$$\Sigma_{i=1}^n i^{1+\frac{1}{i}}$$.

There is no simple formula so you can't just plug a number in.

I recommend that you try some values of n and see what happens:

If n= 1, the sum is simply $1^{1+1}= 1$
If n= 2, the sum is $1+ 2^{1+1/2}= 1+ 2\sqrt{2}$
If n= 3, the sum is $1+ 2\sqrt{2}+ 3^{1+ 1/3}= 1+ 2\sqrt{2}+ 3(3)^{\frac{1}{3}}$

I see a pattern but I don't see any simple way of writing that.

 

[M98Ranger]

...

Hi there!

Finding the partial sum of a series can be a tricky task, but don't worry, I'm here to help. In order to find the partial sum of the series you mentioned, you can use the formula for the partial sum of a geometric series:

Sn = a(1-r^n)/(1-r)

In this case, a = 1 and r = 1/n. Plugging these values into the formula, we get:

Sn = 1(1-(1/n)^n)/(1-(1/n))

Simplifying this expression, we get:

Sn = (1-(1/n)^n)/(1/n-1)

Now, as n approaches infinity, (1/n)^n becomes 0 and 1/(1/n-1) becomes 1. Therefore, the partial sum of the series is 1. This means that as n gets larger and larger, the partial sum gets closer and closer to 1, but it never reaches 1.

I hope this helps you understand how to find the partial sum of a series. If you have any further questions, feel free to ask!