WOLFRAM|DEMONSTRATIONS PROJECT

Sum of the Alternating Harmonic Series (II)

​
n
3
The striped yellow area is
2n
∑
k=n+1
1
k
, so
2n
∑
k=1
k+1
(-1)
1
k
=
2n
∑
k=1
1
k
-2
n
∑
k=1
1
2k
=
2n
∑
k=n+1
1
k
<ln2
.
The striped blue area is
2n
∑
k=n+1
1
k-1
=
2n-1
∑
k=n
1
k
, so
2n-1
∑
k=1
k+1
(-1)
1
k
=
2n-1
∑
k=1
1
k
-2
n-1
∑
k=1
1
2k
=
2n-1
∑
k=n
1
k
>ln2
.
So
2n
∑
k=1
k+1
(-1)
1
k
<ln2<
2n-1
∑
k=1
k+1
(-1)
1
k
, and therefore
lim
n∞
2n
∑
k=1
k+1
(-1)
1
k
=
∞
∑
k=1
k+1
(-1)
1
k
=ln2
.