Partial Sum Approximation for Alternating Harmonic Series

In summary: Sure. I was not criticizing your post as it was very informative! I was merely informing the OP of a more elementary method.
  • #1
DavidE721
1
0

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.
 
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  • #2
DavidE721 said:

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
[tex] \Psi(x) = \frac{d}{dx} \ln \left( \Gamma(x) \right)
= \frac{\Gamma^{\prime}(x)}{\Gamma(x)}.[/tex] 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,
[tex] \sum_{k=1}^N (-1)^{k-1} \frac{1}{k}
= \ln(2) + \frac{1}{2}(-1)^N \left[ \Psi\left(\frac{N}{2}+\frac{1}{2}\right) -
\Psi\left(\frac{N}{2}+1\right) \right].[/tex]
 
  • #3
I doubt you know digamma functions, so that's not going to be the way to solve it.

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

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

What can you say about [itex]|s_{2n+1}-s_{2n}|[/itex]?? What does this say about [itex]|s-s_n|[/itex]?? Do you obtain a formula for how well the approximation is?
 
  • #4
micromass said:
I doubt you know digamma functions, so that's not going to be the way to solve it.

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

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

What can you say about [itex]|s_{2n+1}-s_{2n}|[/itex]?? What does this say about [itex]|s-s_n|[/itex]?? 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).
 
  • #5
Ray Vickson said:
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.
 

Related to Partial Sum Approximation for Alternating Harmonic Series

1. What is Partial Sum Approximation for Alternating Harmonic Series?

Partial Sum Approximation for Alternating Harmonic Series is a mathematical method for estimating the sum of an alternating harmonic series, which is a series in which the signs of the terms alternate between positive and negative.

2. How does Partial Sum Approximation for Alternating Harmonic Series work?

The method involves calculating the sum of a certain number of terms in the series and using that as an estimate for the full sum. The more terms that are included, the closer the approximation will be to the actual sum of the series.

3. Why is Partial Sum Approximation for Alternating Harmonic Series useful?

This method is useful because it allows us to estimate the sum of an alternating harmonic series without having to calculate each individual term, which can be tedious and time-consuming for larger series.

4. What is the formula for Partial Sum Approximation for Alternating Harmonic Series?

The formula is Sn = 1 - (1/2) + (1/3) - (1/4) + ... + (-1)n(1/n), where n is the number of terms included in the partial sum.

5. How accurate is Partial Sum Approximation for Alternating Harmonic Series?

The accuracy of the approximation depends on the number of terms included in the partial sum. The more terms included, the closer the approximation will be to the actual sum of the series. However, it will never be an exact value.

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