Riemann Sums (1 Viewer)

leehuan · May 5, 2016

What does this question actually mean?

$$Find the limits of the sequences whose $n$th terms are:\\ a)$ \frac { {1}^{ a }+ {2}^{ a }+\dots + {n}^{ a } }{ {n}^{ 1+a } }\quad (a>1)\\$b) $\frac{1}{2n+1}+\frac{1}{2n+2}+\dots+\frac{1}{3n}$

seanieg89 · May 5, 2016

It means find

$\lim_{n\rightarrow\infty} T_n$

where T_n is the given expression in the two parts.

Paradoxica · May 5, 2016

leehuan said:
What does this question actually mean?

$$Find the limits of the sequences whose $n$th terms are:\\ a)$ \frac { {1}^{ a }+ {2}^{ a }+\dots + {n}^{ a } }{ {n}^{ 1+a } }\quad (a>1)\\$b) $\frac{1}{2n+1}+\frac{1}{2n+2}+\dots+\frac{1}{3n}$

The first sequence is the integral from 0 to 1 of x^a dx

The second sequence is the integral from 2 to 3 of x^-1 dx

leehuan · May 5, 2016

Paradoxica said:
The first sequence is the integral from 0 to 1 of x^a dx

The second sequence is the integral from 2 to 3 of x^-1 dx

Zzz thanks for curing my blindness earlier in the day.
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$$Find $\int_1^2{x^{-2}dx}$ by using Riemann sums. Use a uniform partition but set $x_i^*=\sqrt{x_{i-1} x_i}$, and use $ \frac{1}{c(c+1)}=\frac{1}{c}-\frac{1}{c+1}$

seanieg89 · May 5, 2016

$I=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{k=1}^n f(\sqrt{x_{k-1}x_k})\\ \\ = \lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^n \frac{n^2}{(n+k)(n+k-1)}\\ \\ = \lim_{n\rightarrow\infty} n\sum_{k=1}^n \left(\frac{1}{n+k-1}-\frac{1}{n+k}\right)\\ \\ = \lim_{n\rightarrow\infty} n(\frac{1}{n}-\frac{1}{2n})=\frac{1}{2}.$

where the first equality comes from the definition of the Riemann sum. (And the fact the the geometric mean of two non-negative numbers lies between them, so we are indeed considering valid tagged partitions.)

Riemann Sums (1 Viewer)

leehuan

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seanieg89

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Paradoxica

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