Oh I see, you are not familiar with the method of differences?
Ok, I show the working. First, we calculate the sum of the first N
terms. (Sum from 1 to N.) After that, we see what happens when
N → ∞.
N
Σ ln (1+1/x)
x=1
= N
= Σ ln (x+1)/x
= x=1
= N
= Σ ln(x+1) − ln
x
= x=1
= (ln2−ln1) + (ln3−ln2)
+ (ln4−ln3)
+ (ln5−ln4) + ...
+ (ln(N+1)−lnN).
= ln(N+1) − ln1.
Because all the positive and negative terms in
between cancel out. This is the method of differences. You can
often use this method to calculate the sum of the first N terms.
It's quite important at A-levels actually. Always come
out.
= ln(N+1).
So we have shown that:
N
Σ ln(1+1/x) =
ln(N+1). Sum of the
first N terms.
x=1
Now we see what happens as N → ∞. If you are familiar with the
ln function, it is quite obvious that, as N → ∞, ln(N+1) → ∞. So:
∞
Σ ln(1+1/x)
= ∞.
x=1
