Infinite Series Part II
We've seen that an infinite sum can sometimes fail to exist for "obvious" reasons. Examples 2, 4, and 5 are more subtle.
Let's look at examples 2 and 4: can these sums exist? Note that one does have an infinite number of terms in each sum. But also note that each term is getting progressively smaller. So, there might be a chance that these sums exist!
It turns out that the sum of example 4 indeed exists, but the sum of example 2 does not. Here is how we can see this:
In example 1, recall from calculus the idea of an indefinite integral. Integrate the function "1/x" from x = 1 to x = "3". The value of this integral is "ln(3)". Notice what "1 + 1/2" represents: it represents the left hand Riemann sum with n = 2 and delta x = 1 for this integral and remember that because the function 1/x is decreasing, the left hand sum is greater than the integral.
Now keep doing this; note that 1+ 1/2 + ....+1/n represents the left hand sum with delta x = 1, n = n for the function 1/x. Notice that this partial sum exceeds the value of the integral of 1/x as x goes from 1 to n; that is, 1+1/2 +....1/n exceeds ln(n).
So we see that the n'th partial sum exceeds ln(n) and we know from calculus that ln(n) increases without bound as n increases. Hence these partial sums will eventually exceed and finite number. This sum cannot make sense.
On the other hand, if 1 + 1/2 + ....+(1/2)^n = Sn
Then we can multiply both sides by "1/2" to get
1/2 + 1/4 + ....(1/2)^(n+1) = (1/2)*Sn
Now subtract: Sn - (1/2)Sn = 1 - (1/2)^(n+1) (subtract the two sums and note that the leading 1 doesn't get subtracted off and the last term (1/2)^(n+1) isn't cancelled either.)
Now solve for Sn: Sn = (1 - (1/2)^(n+1))/(1 - (1/2))
Now notice that as n gets bigger (as we take a limit as n goes to infinity) the right hand side becomes 1/(1-(1/2))= 1/(1/2) = 2 because (1/2)^(n+1) goes to zero as n goes to infinity.
So, you see that a sequence of partial sums is determined by the limit as n goes to infinity! Here, that limit is 2 and that is the value of this infinite sum.
Example 5 is trickier. It turns out that this sum exists, but we need some theory of sequences to see this. Also example 5 is just downright weird.
Remember back in algebra where you were told that addition is commutative; that is, you could add things in different order and still get the same answer? This is false with infinite sums! To see why, keep reading.
Let's look at examples 2 and 4: can these sums exist? Note that one does have an infinite number of terms in each sum. But also note that each term is getting progressively smaller. So, there might be a chance that these sums exist!
It turns out that the sum of example 4 indeed exists, but the sum of example 2 does not. Here is how we can see this:
In example 1, recall from calculus the idea of an indefinite integral. Integrate the function "1/x" from x = 1 to x = "3". The value of this integral is "ln(3)". Notice what "1 + 1/2" represents: it represents the left hand Riemann sum with n = 2 and delta x = 1 for this integral and remember that because the function 1/x is decreasing, the left hand sum is greater than the integral.
Now keep doing this; note that 1+ 1/2 + ....+1/n represents the left hand sum with delta x = 1, n = n for the function 1/x. Notice that this partial sum exceeds the value of the integral of 1/x as x goes from 1 to n; that is, 1+1/2 +....1/n exceeds ln(n).
So we see that the n'th partial sum exceeds ln(n) and we know from calculus that ln(n) increases without bound as n increases. Hence these partial sums will eventually exceed and finite number. This sum cannot make sense.
On the other hand, if 1 + 1/2 + ....+(1/2)^n = Sn
Then we can multiply both sides by "1/2" to get
1/2 + 1/4 + ....(1/2)^(n+1) = (1/2)*Sn
Now subtract: Sn - (1/2)Sn = 1 - (1/2)^(n+1) (subtract the two sums and note that the leading 1 doesn't get subtracted off and the last term (1/2)^(n+1) isn't cancelled either.)
Now solve for Sn: Sn = (1 - (1/2)^(n+1))/(1 - (1/2))
Now notice that as n gets bigger (as we take a limit as n goes to infinity) the right hand side becomes 1/(1-(1/2))= 1/(1/2) = 2 because (1/2)^(n+1) goes to zero as n goes to infinity.
So, you see that a sequence of partial sums is determined by the limit as n goes to infinity! Here, that limit is 2 and that is the value of this infinite sum.
Example 5 is trickier. It turns out that this sum exists, but we need some theory of sequences to see this. Also example 5 is just downright weird.
Remember back in algebra where you were told that addition is commutative; that is, you could add things in different order and still get the same answer? This is false with infinite sums! To see why, keep reading.
posted by Harriet at 9:09 AM
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