Revisions to Value of divergent sum $\sum_{n=0}^\infty (-1)^n n^n$

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edited Sep 24, 2021 at 4:00

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The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ as it can be seen in the following Pari GP v2.14.0 lines

although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ as it can be seen in the following Pari GP v2.14.0 lines

although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

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edited Sep 24, 2021 at 3:30

Jorge Zuniga

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The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x)}$$W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x)}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x))}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

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edited Sep 23, 2021 at 13:25

Jorge Zuniga

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The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine transformand sine integrals of the derivativederivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x)}$ and $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) \frac {W_{0}(x)}{x(1 + W_{0}(x)}\mathrm{d}x = 0.3233674316...$$$-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine transform of the derivative $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x)}$ $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) \frac {W_{0}(x)}{x(1 + W_{0}(x)}\mathrm{d}x = 0.3233674316...$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$

The sum is equal to $$ \int\limits_{0}^{\infty}\frac{\exp(-x)}{1+W_0(x)}\,\mathrm{d}x = 0.7041699604... $$ where $W_0(x)$ is the Lambert-$W$ function.

Reference

Stephen Finch. "Errata and Addenda to Mathematical Constants [Math.HO]", §6.11. Iterated Exponential Constants, pg. 66.

Two interesting divergent sums from this reference are as well $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}e^{-x}\log(x/|W_{0}(i x)|)\mathrm{d}x = 0.3233674316...$$ which seems to be also the cosine and sine integrals of the derivatives $W'_{0}(x) = \frac {W_{0}(x)}{x(1 + W_{0}(x)}$ and $-W''_{0}(x)$ respectively $$\sum_{n=1}^\infty (-1)^{n-1} (2n)^{2n-1} = \int\limits_{0}^{\infty}\mathrm{cos}(x) W'_{0}(x)\mathrm{d}x = -\int\limits_{0}^{\infty}\mathrm{sin}(x) W''_{0}(x)\mathrm{d}x$$ although, as it is pointed out, a rigorous proof is not yet known.

And this variation, $$\sum_{n=1}^\infty (-1)^{n-1} (2n-1)^{2n} = \frac{1}{2i}\int\limits_{0}^{\infty}e^{-x}(g(ix)-g(-ix))\mathrm{d}x = 0.0111203007...$$ where $g(x) = W_{0}(x)/(1 + W_{0}(x))^3$