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Let $b>1$ be a real number and let $f$ be given on the interval $[1,b]$ by $f(x)=x^2$. A partition $P_n$ are defined for every postive whole number $n$ as follows:

$P_n: x_0=1, x_1=b^{\frac{1}{n}}, x_2=b^{\frac{2}{n}}+...+x_{n-1}=b^{\frac{n-1}{n}}, x_n=b^{\frac{n}{n}}=b $.

Show that the upper- and lower Riemann sum is given by:

$U(f,P_n)=b^{\frac{2}{n}}(\frac{b^{1/n}-1}{b^{3/n}-1})(b^3-1)$ and $L(f,P_n)=(\frac{b^{1/n}-1}{b^{3/n}-1})(b^3-1)$

Hint: Use the formula $\sum_{i=1}^{n}b^{3(i-1)/n}=\frac{b^{3}-1}{b^{3/n}-1}$

What I have tried so far:

$U(f,P_n)=f(u_1)\Delta x_1+f(u_2)\Delta x_2+f(u_3)\Delta x_3+...+$ $=1^2(b^{1/n}-1)+b^{2/n}(b^{2/n}-b^{1/n})+b^{4/n}(b^{3/n}-b^{2/n})+b^{6/n}(b^{4/n}-b^{3/n})+...+b^{\frac{2n-2}{n}}(b-b^{\frac{n-1}{n}})$

$=b^{1/n}-1+b^{4/n}-b^{3/n}+b^{7/n}-b^{6/n}+b^{10/n}-b^{9/n}+...+b^{3n-2/n}-b^{3n-3/n}$

From here, I am not able to see any "obvious" traits on the sequence

Mr. P
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  • And... what comes next? – Did Mar 29 '16 at 21:52
  • I answered this question here: http://math.stackexchange.com/questions/1716734/show-that-for-any-epsilon-there-exists-n-so-that-uf-p-n-lf-p-n-epsilo/1716745#1716745 – User8128 Mar 29 '16 at 22:00
  • @ User8128: The starting point is the same, but the question is different... – Mr. P Mar 29 '16 at 22:14
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    @Did: I've edited the post showing by showing the next step by multiplying out the parentheses but I am still not able to see any "obvious" traits – Mr. P Mar 29 '16 at 22:17
  • Hmmm... for example, $$b^{1/n}+b^{4/n}+b^{7/n}+b^{10/n}+\cdots+b^{3-2/n}=b^{1/n}\sum_{k=0}^{n-1}a^k$$ for some well chosen $a$, and the sum in the RHS is... hence the LHS is... Already half of the work is done. Then, the other terms are... hence... – Did Mar 30 '16 at 18:26

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