In this video, I will present an example
of how to calculate the value of a series directly from the definition.
It will be this one.
I invite you to pause the video and try to calculate it,
and then continue watching.
I want to calculate the sum from n = 1 to ∞, of 1/(n² + n).
By definition of series,
what I have to do is find a formula for the sum of the first k terms,
and then take the limit as k → ∞.
And if I don't know what else to try,
it won't hurt to simply compute the first few of those partial sums,
and see what we get from there. Let's do that!
The first partial sum S₁, is simply the first term: 1/(1² + 1), and that is 1/2.
The second partial sum S₂, is the first partial sum + the second term.
This is 1/2 + 1/6, and this comes up to be 2/3.
The third partial sum is the second partial sum + the third term.
(when n = 3) That is 1/(3² + 3),
so that is 2/3 + 1/12, and this simplifies to 3/4;
and I'm going to stop here, because I think I already see a pattern.
S₁ = 1/2, S₂ = 2/3, S₃ = 3/4... I think I can guess a formula here.
This is just a guess so far so I'm going to call it a conjecture,
but it appears that for every k ≥ 1, the k-th partial sum is k/(k + 1).
This certainly works for the first three.
This of course needs to be proven, we cannot simply state it.
And perhaps the simplest way to do this proof is by induction,
because from the definition,
the (k + 1)-st partial sum can be written as the k-th partial sum + a term,
so if I have a formula for the k-th partial sum,
I can easily get a formula for the (k + 1)-st partial sum.
This is a standard proof by induction, and it works fine.
I won't write down the details.
However, I'll notice that to do it this way,
I need to be able to guess the correct formula.
There is an alternative approach for this problem,
even if you are not able to guess the correct formula.
Here is a different way to obtain the same result without any guessing.
I want a formula for the k-th partial sum,
which is the sum from n = 1 to k, of 1/(n² + n).
And I notice first that this denominator can be factored,
so I'm going to rewrite it as 1/(n⋅(n + 1)),
and this makes me think of partial fraction decomposition.
I can split this as the sum of two fractions with simpler denominators.
(In one case n, and in the other case (n + 1).)
You can use a more systematic way here,
but I'm just going to do a little bit of guess-and-check.
I think if we put a 1 here and a 1 here, and I subtract it works,
because the denominator will be n⋅(n + 1),
and the numerator is (n + 1) - n, which is 1.
Yes, this works exactly!
So, I have simply rewritten this algebraically.
And now, because I am working with a finite sum,
I can use all the manipulations we are used to using with partial sums.
I can use properties, and regroup terms, and all that...
So I'm going to do exactly that.
I have k terms here (every term is the difference of two).
I'm going to expand and see if some of the terms cancel, and they will.
So when n = 1, I get 1 - 1/2, when n = 2, I get 1/2 - 1/3,
and look at what's happening here:
I already have a cancellation between this 1/2, and this 1/2.
When n = 3, I get + 1/3 - 1/4,
and again, I already have a cancellation between this 1/3 and this 1/3.
And it looks like almost all the terms are going to cancel.
So I keep doing this...
The last term will be when n = k, and I get 1/k - 1/(k + 1).
And again, all the terms cancel this - 1/4 cancels with something there,
this 1/k will cancel with - 1/k, and I'm left with only two terms:
the first term, which was 1, and the last term, which is - 1/(k + 1),
and this simplifies as k/(k + 1).
This is the formula for Sₖ that I obtained before (exactly the same one),
except that now I didn't have to do any guessing.
And also, this by itself is a proof!
If I derive this formula this way, it's already proven.
I don't have to use induction or anything else next.
This is good enough by itself.
So either in the previous way,
(I guessed the formula and proved it by induction)
or in this way (using this manipulation),
I know what the sum of the first k terms is;
therefore, I can now use the definition of series,
to calculate the value of the sum with infinitely many terms.
I know that the original sum (sum of 1/(n² + n), from n = 1 to ∞),
is simply the limit as k → ∞, of the sum of the first k terms;
and since I now have a formula for that, I can compute this limit.
It's the limit as k → ∞, of k/(k + 1), and that is simply 1.
So we have concluded, directly from the definition,
that this series is convergent and that it adds up to 1.