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Another divergence proof

Here is another proof for the divergence for the harmonic series . I assume no one has found it earlier! (its awfully dumb to say I'm the first one!!)..... consider the integral \[ I(a,x) = \int_0^{x} \frac{(\sin x)^{a}}{\cos x} dx \] we will tease this until it cries out an interesting series .....

multiplying 1 continuously OR \[ (\sin x)^2 + (\cos x)^2 \] yields \[ = \int_0^{x} (\frac{(\sin x)^{a}}{\cos x})((\sin x)^2 + (\cos x)^2) ...... dx \] \[ = \int_0^{x} (\sin x)^{a}\cos x + (\frac{(\sin x)^{a+2}}{\cos x})((\sin x)^2 + (\cos x)^2) dx \] \[ = \int_0^{x} ( (\sin x)^{a}\cos x + (\sin x)^{a+2}\cos x + (\sin x)^{a+4}\cos x ....... \infty ) dx \] \[ = ( \frac{(\sin x)^{a+1}}{a+1} + \frac{(\sin x)^{a+3}}{a+3} + \frac{(\sin x)^{a+5}}{a+5} ..... \infty ) \] plug in \[ a=1 ; x = \frac{\pi}{2} \] \[ = \frac{1}{a+1} + \frac{1}{a+3} + \frac{1}{a+5} ...... \infty \] to get \[ = \frac{1}{2} ( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} ...... \infty ) \] which is half the harmonic series ........ solving the integral for the same a=1 gives \[ I(1,\frac{\pi}{2}) = \int_0^{\frac{\pi}{2}} \tan x dx \] \[ = \infty \] ....... hence the harmonic series diverges!

What is more important here is the behavior of the series as the limit transforms from 0 to x ..... When x starts from 0 further (keeping a as 1 ) and reaches pi/2 the series changes its values according to the function tan x ....... beautiful! isn't it ?

please post your comments and ideas!

Note by Abhinav Raichur
1 year, 11 months ago

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