The method I used is recognizing the limit is actually a definite integral (in the form of a Riemann sum). (I wouldn't be surprised if someone else posted an easier solution involving a subtle algebraic manipulation of the actual limit instead of converting it to an integral).
Maybe you remember that a definite integral is defined as a Riemann sum. If not, then here is the definition of a definite integral:
For the rest of this post, I am going to show that your problem is actually a Riemann sum (the right-hand side of the above equation). By doing so, I am allowed to convert it to an equivalent definite integral, which will hopefully be easy to evaluate!
If the question were to find the area between f(x) = x^199 and the x-axis on the interval [a,b] using a Riemann sum with n rectangles, what would you do? The Riemann sum would look like:
If we decide each of our rectangles will have the same width, then
And then, .
Substituting these values into the Riemann sum gives:
This looks nothing like the original problem, right? Well, the first term of our summation is (i = 1):
If we look back at the original problem, we can imagine the first term being , which is equal to . They are starting to look similar (a little!).
Now, the natural question is what values of a and b will make the first term in the Riemann sum and the first term in the original problem match exactly?
You may already notice that will probably have to match . Similarly, will probably have to match . Let's focus on and .
doesn't have any addition. It isn't , it's just . If a = 0, then becomes . They match!
Now, let's switch our focus to and . We now know a = 0, so:
will have to match .
We can conclude b = 1. Substituting a and b into the Riemann sum and checking other terms, such as i = 2 or i = 3, shows that our Riemann sum now matches the original problem exactly. In other words, I have just shown that your problem is actually a Riemann sum with a = 0 and b = 1. According to the definition of a definite integral, a Riemann sum is just a definite integral. Therefore, your problem is a definite integral!
Now we know the original problem is equivalent to evaluating , where a = 0 and b = 1.
Like I said, I wouldn't be surprised if someone now posts a much more elegant solution that has eluded me. Until then, this is the best I've got.
In summary, the original problem is actually a definite integral. Once the limits of integration were found, it was only a matter of evaluating an integral.
Hi, Jose. <3