Partial Sum Approximation for Alternating Harmonic Series

[DavidE721]

Homework Statement

Find a value for n for which the nth partial sum is ensured to approximate the sum of the alternating harmonic infinite series to three decimal places.

Homework Equations

Sn = Ʃ(-1)^k+1*1/k = 1 - 1/2 + 1/3 - 1/4 + 1/5 - . . .

S1 = 1
S2 = 1 - 1/2
S3 = 1 - 1/2 + 1/3
S4 = 1 - 1/2 + 1/3 -1/4

.
.
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Sn = ?

The Attempt at a Solution

An attempt was made to derive a formula that would permit finding the value of the nth partial sum (e.g., S1000 = ?), but without success. It appears as though such a formula might not exist. Any help in this regard would be most appreciated.

 

[Ray Vickson]

Homework Statement

Find a value for n for which the nth partial sum is ensured to approximate the sum of the alternating harmonic infinite series to three decimal places.

Homework Equations

Sn = Ʃ(-1)^k+1*1/k = 1 - 1/2 + 1/3 - 1/4 + 1/5 - . . .

S1 = 1
S2 = 1 - 1/2
S3 = 1 - 1/2 + 1/3
S4 = 1 - 1/2 + 1/3 -1/4

.
.
.

Sn = ?

The Attempt at a Solution

An attempt was made to derive a formula that would permit finding the value of the nth partial sum (e.g., S1000 = ?), but without success. It appears as though such a formula might not exist. Any help in this regard would be most appreciated.

There is a formula, but it involves the non-elementary function "digamma" function Psi(x), defined as
$$\Psi(x) = \frac{d}{dx} \ln \left( \Gamma(x) \right) <br /> = \frac{\Gamma^{\prime}(x)}{\Gamma(x)}.$$ Here, $\Gamma$ is the standard Gamma function. According to Maple 11 the sum is:
Sum((-1)^(k-1)/k,k=1..N):lprint(%);Sum((-1)^(k-1)/k,k = 1 .. N) <--echo the input

S:=value(%): lprint(S);
ln(2)+1/2*(-1)^(N+1)*Psi(1+1/2*N)+1/2*(-1)^N*Psi(1/2*N+1/2). <--- output

That is,
$$\sum_{k=1}^N (-1)^{k-1} \frac{1}{k}<br /> = \ln(2) + \frac{1}{2}(-1)^N \left[ \Psi\left(\frac{N}{2}+\frac{1}{2}\right) -<br /> \Psi\left(\frac{N}{2}+1\right) \right].$$

 

[micromass]

I doubt you know digamma functions, so that's not going to be the way to solve it.

Let $s=\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n}$ and let $s_n$ be the n-th partial sum.

Can you show that $s_{2n+1}$ is decreasing? And that $s_{2n}$ is increasing? Can you conclude from that that $s_{2n}\leq s \leq s_{2n+1}$?

What can you say about $|s_{2n+1}-s_{2n}|$?? What does this say about $|s-s_n|$?? Do you obtain a formula for how well the approximation is?

 

[Ray Vickson]

I doubt you know digamma functions, so that's not going to be the way to solve it.

Let $s=\sum_{n=1}^{+\infty} \frac{(-1)^{n+1}}{n}$ and let $s_n$ be the n-th partial sum.

Can you show that $s_{2n+1}$ is decreasing? And that $s_{2n}$ is increasing? Can you conclude from that that $s_{2n}\leq s \leq s_{2n+1}$?

What can you say about $|s_{2n+1}-s_{2n}|$?? What does this say about $|s-s_n|$?? Do you obtain a formula for how well the approximation is?

I agree about it not being the way to solve it, but he/she did ask. If one is going to use a computer, one might as well just do a recursive computation (although avoiding subtractive roundoff errors is an issue for getting accurate numbers).

 

[micromass]

I agree about it not being the way to solve it, but he/she did ask. If one is going to use a computer, one might as well just do a recursive computation (although avoiding subtractive roundoff errors is an issue for getting accurate numbers).

Sure. I was not criticizing your post as it was very informative! I was merely informing the OP of a more elementary method.