Find the infinite sum of fractions 2/(3⋅5)+(2⋅4)/(3⋅5⋅7)+(2⋅4⋅6)/(3⋅5⋅7⋅9)+....

[lfdahl]

I´m not sure, whether this little challenge has been posted before. I have searched the forum and didn´t find it.
It might still be a duplicate though ...
Find the sum of fractions

$$\frac{2}{3\cdot5}+\frac{2\cdot4}{3\cdot5\cdot7}+\frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9}+...$$

 

[tkhunny]

I´m not sure, whether this little challenge has been posted before. I have searched the forum and didn´t find it.
It might still be a duplicate though ...
Find the sum of fractions

$$\frac{2}{3\cdot5}+\frac{2\cdot4}{3\cdot5\cdot7}+\frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9}+...$$

I can't figure out how to use the HIDE feature, so I won't answer, but this one popped out immediately. Maybe I have seen it before, too. I don't recall anywhere that may have happened.

 

[Opalg]

I can't figure out how to use the HIDE feature

Use the tags [sp] ... [/sp] to hide spoilers.

 

[tkhunny]

Use the tags [sp] ... [/sp] to hide spoilers.

"SPOILERS"! Too easy. Many thanks.

All right, then...

[sp]With some judicious factoring, rewrite as

$\dfrac{2}{3}\cdot\left(\dfrac{1}{5} + \dfrac{4}{5}\cdot\left(\dfrac{1}{7} + \dfrac{6}{7}\cdot\left(\dfrac{1}{9}+\dfrac{8}{9}\cdot\left(\dfrac{1}{11}+\dfrac{10}{11}\cdot\left(...\right)\right)\right)\right)\right)$

Pretty clearly, we have 2/3 and everything else is a REALLY long-winded way to say 1.[/sp]

 

[tkhunny]

Less sure the more I look at it. See, that's why I don't usually answer these things. They invade my brain if I let them.

 

[lfdahl]

Less sure the more I look at it. See, that's why I don't usually answer these things. They invade my brain if I let them.

Your solution is correct, as far as I can see. (Yes) Thankyou, tkhunny for your participation!(Sun)

Yes, I know this feeling too, it might sometimes occupy your mind both night and day ...(Whew)

 

[Opalg]

[sp]With some judicious factoring, rewrite as

$\dfrac{2}{3}\cdot\left(\dfrac{1}{5} + \dfrac{4}{5}\cdot\left(\dfrac{1}{7} + \dfrac{6}{7}\cdot\left(\dfrac{1}{9}+\dfrac{8}{9}\cdot\left(\dfrac{1}{11}+\dfrac{10}{11}\cdot\left(...\right)\right)\right)\right)\right)$

Pretty clearly, we have 2/3 and everything else is a REALLY long-winded way to say 1.[/sp]

[sp]That argument leaves me feeling a bit queasy, but I believe it gives the correct answer. Here's another approach.

Form the power series $$y = \frac{1}{3}x^3 +\frac{2}{3\cdot5}x^5 +\frac{2\cdot4}{3\cdot5\cdot7}x^7 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9}x^9 + \ldots.$$

Differentiate: $$y' = x^2 + \frac{2}{3}x^4 +\frac{2\cdot4}{3\cdot5}x^6 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^8 + \ldots.$$

Then $$\begin{aligned} \frac{y' - x^2}{x^3} &= \frac13(2x) + \frac{2}{3\cdot5}(4x^3) + \frac{2\cdot4}{3\cdot5\cdot7}(6x^5) + \ldots \\ &= \frac{d}{dx}\left(\frac yx\right) \\ &= \frac{xy'-y}{x^2}, \\ y' - x^2 &= x^2y'- xy, \\ y'(1-x^2) + xy &= x^2.\end{aligned}$$
Solve that differential equation by using an integrating factor $(1-x^2)^{-3/2}$, to get $y'(1-x^2)^{-1/2} + x(1-x^2)^{-3/2}y = x^2(1-x^2)^{-3/2}.$ That integrates to give $$\frac y{(1-x^2)^{1/2}} = \int\frac{x^2}{(1-x^2)^{3/2}}dx = \int \tan^2\theta\, d\theta = \int( \sec^2\theta - 1)\,d\theta = \tan\theta - \theta = \frac x{\sqrt{1-x^2}} - \arcsin x,$$ using the substitution $x = \sin\theta$ and choosing the constant of integration so that $y=0$ when $x=0$.

So $y = x - \sqrt{1-x^2}\arcsin x.$ When $x=1$, $y=1$. But when $x=1$ the power series for $y$ becomes $$\frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots.$$ This shows that $$1 = \frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots$$, and subtracting $\dfrac13$ from both sides gives the required answer.

[/sp]

 

[tkhunny]

[sp]That argument leaves me feeling a bit queasy, but I believe it gives the correct answer. Here's another approach.

Agreed. It may be dumb luck and confirmation bias.

 

[lfdahl]

[sp]That argument leaves me feeling a bit queasy, but I believe it gives the correct answer. Here's another approach.

Form the power series $$y = \frac{1}{3}x^3 +\frac{2}{3\cdot5}x^5 +\frac{2\cdot4}{3\cdot5\cdot7}x^7 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9}x^9 + \ldots.$$

Differentiate: $$y' = x^2 + \frac{2}{3}x^4 +\frac{2\cdot4}{3\cdot5}x^6 + \frac{2\cdot4\cdot6}{3\cdot5\cdot7}x^8 + \ldots.$$

Then $$\begin{aligned} \frac{y' - x^2}{x^3} &= \frac13(2x) + \frac{2}{3\cdot5}(4x^3) + \frac{2\cdot4}{3\cdot5\cdot7}(6x^5) + \ldots \\ &= \frac{d}{dx}\left(\frac yx\right) \\ &= \frac{xy'-y}{x^2}, \\ y' - x^2 &= x^2y'- xy, \\ y'(1-x^2) + xy &= x^2.\end{aligned}$$
Solve that differential equation by using an integrating factor $(1-x^2)^{-3/2}$, to get $y'(1-x^2)^{-1/2} + x(1-x^2)^{-3/2}y = x^2(1-x^2)^{-3/2}.$ That integrates to give $$\frac y{(1-x^2)^{1/2}} = \int\frac{x^2}{(1-x^2)^{3/2}}dx = \int \tan^2\theta\, d\theta = \int( \sec^2\theta - 1)\,d\theta = \tan\theta - \theta = \frac x{\sqrt{1-x^2}} - \arcsin x,$$ using the substitution $x = \sin\theta$ and choosing the constant of integration so that $y=0$ when $x=0$.

So $y = x - \sqrt{1-x^2}\arcsin x.$ When $x=1$, $y=1$. But when $x=1$ the power series for $y$ becomes $$\frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots.$$ This shows that $$1 = \frac{1}{3} +\frac{2}{3\cdot5} +\frac{2\cdot4}{3\cdot5\cdot7} + \frac{2\cdot4\cdot6}{3\cdot5\cdot7\cdot9} + \ldots$$, and subtracting $\dfrac13$ from both sides gives the required answer.

[/sp]

Thankyou for your participation, Opalg! I wonder, how you get these ideas. A fascinating approach indeed!(Wink)