Message Boards Message Boards

GROUPS:

Precision of numerical series value in Wolfram|Alpha

Posted 2 months ago
442 Views
|
2 Replies
|
0 Total Likes
|

Error

It is gives $$\sum_{k=0}^{\infty}\frac{(-1)^k}{k^2+2k+2}\approx 0.363998$$ but exact value: $$\sum_{k=0}^{\infty}\frac{(-1)^k}{k^2+2k+2}=\frac{1}{2}-\frac{\pi}{e^{\pi}-e^{-\pi}}\approx 0.3639854725089334$$

POSTED BY: Yuri Artamonov
2 Replies

With Mathematica 13.1,numerically:

 NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}, 
  Method -> "AlternatingSigns", WorkingPrecision -> 32]
(*0.36398547250893341852488170865923*)

NSum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}, 
 Method -> "WynnEpsilon", NSumTerms -> 10^5, WorkingPrecision -> 32]
(*0.36398547250893341852488170816398*)

Symbolically:

Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}]
(*1/2 (1 - \[Pi] Csch[\[Pi]])*)
N[%, 32]
(*0.36398547250893341852488170865923*)

If you use WolframAlfa answer is not precisie ,because for quick answer to you from web. Try to put:

N[Sum[(-1)^k/(k^2 + 2 k + 2), {k, 0, Infinity}] ,32]

Give you: 0.36398547250893341852488170816398

POSTED BY: Mariusz Iwaniuk
Posted 2 months ago

Thanks. Yes, I used Wolframa|Alpha. But I think that it is not a good idea to have more velocity and error results from web form.

POSTED BY: Yuri Artamonov
Reply to this discussion
Community posts can be styled and formatted using the Markdown syntax.
Reply Preview
Attachments
Remove
or Discard

Group Abstract Group Abstract