Analysis & Series Homework #1 & #2

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Homework Statement


Number 1.

an=(1+1/n)n for all n
bn=(1-1/n)-n n=2,3,4,...

i) Prove that 1<bn/an<=n/(n-1) n=2,3,4...
ii)Show that <bn> also converges to e'. "e" being the exponential.

Number 2.

i) Suppose that 1<= r <= n for all r and all n. Prove 1 + series from m=1 to r [(1/m!)*(1-1/n)*(1-2/n)...(1-(m-1)/n)] <= an <= e.

ii) Prove that 1 + series from m=1 to r [1/m!] <= e' <=e.

iii) prove that e'=e.number 2 I understand conceptually it makes logical sense but I am not sure how to prove it.
 
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erikalculator said:

Homework Statement


Number 1.

an=(1+1/n)n for all n
bn=(1-1/n)-n n=2,3,4,...

i) Prove that 1<bn/an<=n/(n-1) n=2,3,4...
ii)Show that <bn> also converges to e'. "e" being the exponential.

Number 2.

i) Suppose that 1<= r <= n for all r and all n. Prove 1 + series from m=1 to r [(1/m!)*(1-1/n)*(1-2/n)...(1-(m-1)/n)] <= an <= e.

ii) Prove that 1 + series from m=1 to r [1/m!] <= e' <=e.

iii) prove that e'=e.


number 2 I understand conceptually it makes logical sense but I am not sure how to prove it.

Maybe try using the conjugate of the denominator and multiplying that by the top and bottom of your fraction. I think that could help, but I also think that there may be more information needed to solve this. :(
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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