Series to Ad Infinitum part III

Calculus Level 4

1 1 ⋅ 2 − 1 2 ⋅ 3 + 1 3 ⋅ 4 − ⋯ = ? \displaystyle{ \dfrac{1}{1\cdot 2} -\dfrac{1}{2\cdot 3} +\dfrac{1}{3\cdot 4} - \cdots = \text{ ?} }

Answer up to 4 4 decimal places.

Image Credit: Wikipedia Natural Logarithm by W. Muła


The answer is 0.3862.

Kunal Gupta by Kunal Gupta , 23, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

3 solutions

This series can be written as

∑ n = 1 ∞ ( − 1 ) n + 1 ( 1 n − 1 n + 1 ) = \displaystyle\sum_{n=1}^{\infty} (-1)^{n+1} \left( \dfrac{1}{n} - \dfrac{1}{n+1} \right) =

1 + 2 ∗ ∑ n = 2 ∞ ( − 1 ) n + 1 n = 1 + 2*\displaystyle\sum_{n=2}^{\infty} \dfrac{(-1)^{n+1}}{n} =

1 + 2 ∗ ( ln ⁡ ( 2 ) − 1 ) = 2 ln ⁡ ( 2 ) − 1 = 0.386 1 + 2*(\ln(2) - 1) = 2\ln(2) - 1 = \boxed{0.386} to 3 decimal places.

Note that the alternating harmonic series ∑ n = 1 ∞ ( − 1 ) n + 1 n = ln ⁡ ( 2 ) . \displaystyle\sum_{n=1}^{\infty} \dfrac{(-1)^{n+1}}{n} = \ln(2).

10 Helpful 0 Interesting 0 Brilliant 1 Confused

Sir , I'am just little extending your solution , because i enjoyed in solving this question. So I'am just sharing it. :)

S = ∑ r = 1 ∞ ∫ 0 − 1 x r r d x S = − ∑ r = 1 ∞ ∫ − 1 0 x r r d x S = − ∫ − 1 0 ( ∑ r = 1 ∞ x r r ) d x y = ∑ r = 1 ∞ x r r d y d x = ∑ r = 1 ∞ x r − 1 = 1 + x + x 2 + . . . . . . ( G P ) d y d x = 1 1 − x \displaystyle{{ S=\sum _{ r=1 }^{ \infty }{ \int _{ 0 }^{ -1 }{ \cfrac { { x }^{ r } }{ r } dx } } \\ S=-\sum _{ r=1 }^{ \infty }{ \int _{ -1 }^{ 0 }{ \cfrac { { x }^{ r } }{ r } dx } } \\ S=-\int _{ -1 }^{ 0 }{ \left( \sum _{ r=1 }^{ \infty }{ \cfrac { { x }^{ r } }{ r } } \right) dx } \\ y=\sum _{ r=1 }^{ \infty }{ \cfrac { { x }^{ r } }{ r } } \\ \cfrac { dy }{ dx } =\sum _{ r=1 }^{ \infty }{ { x }^{ r-1 } } =1+x+{ x }^{ 2 }+......\quad (GP)\\ \cfrac { dy }{ dx } =\cfrac { 1 }{ 1-x } \\ }}

Now Integrating and remove constant of integration by putting initial values : y = − ln ⁡ ( 1 − x ) S = − ∫ − 1 0 − ln ⁡ ( 1 − x ) d x = ∫ − 1 0 ln ⁡ ( 1 − x ) d x 1 − x = t S = ∫ 1 2 ln ⁡ t d t = [ t ln ⁡ t − t ] 1 2 = ln ⁡ 4 − 1 \displaystyle{{ y=-\ln { (1-x) } \\ S=-\int _{ -1 }^{ 0 }{ -\ln { (1-x) } dx } =\int _{ -1 }^{ 0 }{ \ln { (1-x) } dx } \\ 1-x=t\\ S=\int _{ 1 }^{ 2 }{ \ln { t } dt } ={ \left[ t\ln { t } -t \right] }_{ 1 }^{ 2 }=\ln { 4 } -1 }}

Deepanshu Gupta - 6 years, 3 months ago

Log in to reply

Great work, Deepanshu. I think that d y d x = 1 1 − x ⟹ y = − ln ⁡ ( 1 − x ) + c , \frac{dy}{dx} = \frac{1}{1 - x} \Longrightarrow y = -\ln(1 - x) + c, but the elegance of your proof remains intact. :)

Brian Charlesworth - 6 years, 3 months ago

Log in to reply

thanks Sir , I found my typing mistake , I intially eat a negative sign , But thanks sir Now I have corrected that :)

Deepanshu Gupta - 6 years, 3 months ago

Nicely done (Y)

Krishna Sharma - 6 years, 3 months ago

Perhaps it would be clear to take two fractions at a time and show that you can change it to − 1 + ∑ n = 1 ∞ 2 n − 2 n + 1 -1 + \displaystyle \sum_{n=1}^{\infty} \frac{2}{n}-\frac{2}{n+1} . I think ( − 1 ) n (-1)^{n} coefficients are nice and succinct, but only in the right context.

Jake Lai - 6 years, 3 months ago
Noel Lo
Jun 15, 2015

1 1 ∗ 2 − 1 2 ∗ 3 + 1 3 ∗ 4 + . . . . = 1 − 1 2 − 1 2 + 1 3 + 1 3 − 1 4 + . . . . \frac{1}{1*2} -\frac{1}{2*3} + \frac{1}{3*4} +.... = 1 - \frac{1}{2} - \frac{1}{2} + \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ....

= 2 ( 1 − 1 2 + 1 3 − 1 4 + . . . . ) − 1 = 2 l n 2 − 1 = l n 4 − l n e = l n 4 e = 0.386 = 2(1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} +....) - 1 = 2ln 2 - 1 = ln 4 - ln e = ln \frac{4}{e} = \boxed{0.386}

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Just by looking the series I did remember the approximation of
y = Ln (1+x) = x -x^2 + x^3/3 -x^4/4 +...+(-1)^n-1(x^n/n)

valid for the interval -1<x<=1.

In our case x =1 and coincides with the upper limit, so it will render the solution Ln(1+1)

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...