⧼exchistory⧽
11 exercise(s) shown, 0 hidden
BBy Bot
Nov 03'24
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[/math]
For each of the following sequences [math]\{s_n\}[/math], compute [math]\lim_{n \goesto \infty} s_n[/math] if the limit exists.
- [math]s_n = \frac{n+1}{n-1}[/math], for [math]n = 2,3,\ldots[/math].
- [math]s_n = \frac{2n^2 - 3n + 1}{5n^2 + 7}[/math], for [math]n = 1,2,3,\ldots[/math].
- [math]s_n = 1 + \frac1n[/math], for every positive integer [math]n[/math].
- [math]s_n = \frac{n-2}{\sqrt n}[/math], for every positive integer [math]n[/math].
BBy Bot
Nov 03'24
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[/math]
Let sequences [math]\{a_i\}[/math], [math]\{b_j\}[/math], and [math]\{s_n\}[/math] be defined by
[[math]]
a_i = i^3,
[[/math]]
[[math]]
b_j = j-1,
[[/math]]
[[math]]
s_n = \frac1{n+1}.
[[/math]]
Evaluate
- [math]\sum_{i=1}^4 a_i[/math]
- [math]\sum_{j=-2}^2 b_j[/math]
- [math]\sum_{j=1}^3 (2a_j + 5b_j)[/math]
- [math]\sum_{i=1}^4 \frac{a_i}{i+1}[/math]
- [math]\sum_{i=1}^3 s_i[/math]
- [math]\sum_{j=0}^3 a_jb_j[/math].
BBy Bot
Nov 03'24
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[/math]
Compute
- [math]\sum_{i=1}^5 (2i^2 - 3i + 4)[/math]
- [math]\sum_{j=1}^5 [(j+1)^2 - j^2][/math]
- [math]\sum_{k=0}^3 x^k[/math]
- [math]\sum_{j=1}^4 \frac{x^j}j[/math]
- [math](1-x) \sum_{k=0}^3 x^k[/math].
BBy Bot
Nov 03'24
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[/math]
For any sequence [math]a_0, \ldots, a_n[/math], show that [math]\sum_{k=1}^n (a_k - a_{k-1})[/math] depends only on the first and last terms.
BBy Bot
Nov 03'24
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[/math]
Using the various properties of summation, evaluate
- [math]\sum_{i=1}^n (3i^2 + 2)[/math]
- [math]\sum_{j=1}^n (j^2 - 2j + 1)[/math]
- [math]\sum_{i=1}^n \frac{3i^2 + 2}{n^3}[/math]
- [math]\sum_{i=1}^n \frac{2i-2}{n^2}[/math]
- [math]\sum_{i=0}^n (i^2 + i + 1)[/math].
BBy Bot
Nov 03'24
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[/math]
If [math]f(x) = x^2 - x + 1[/math], find
- [math]\sum_{i=1}^4 f(i)[/math]
- [math]\sum_{i=1}^4 f\left( \frac i4 \right)[/math]
- [math]\sum_{i=1}^n f(i)[/math]
- [math]\sum_{i=0}^n f\left( \frac in \right)[/math].
BBy Bot
Nov 03'24
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[/math]
Compute [math]\lim_{n\goesto\infty} S_n[/math], if the limit exists, for each of the following sequences.
- [math]S_n = \sum_{i=1}^n \frac{3i^2 + 2}{n^3}[/math], for every positive integer [math]n[/math].
- [math]S_n = \sum_{i=1}^n \frac{2i-2}{n^2}[/math], for every positive integer [math]n[/math].
- [math]S_n = \sum_{i=1}^n \frac{i+1}{n^3}[/math], [math]n = 1,2,\ldots[/math].
- [math]S_n = \sum_{j=1}^n \frac{j^2+1}n[/math], [math]n = 1,2,\ldots[/math].
BBy Bot
Nov 03'24
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[/math]
Evaluate
- [math]\lim_{n\goesto\infty} \sum_{i=1}^n \frac i{n^2}[/math]
- [math]\lim_{n\goesto\infty} \sum_{i=1}^n \frac{6i^2 - 2i + 1}{n^3}[/math].
BBy Bot
Nov 03'24
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[/math]
Prove \ref{thm 4.2.5} by induction on [math]n[/math].
BBy Bot
Nov 03'24
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[/math]
Using the identity
[[math]]
\sum_{i=1}^n i^3 = \left( \frac{n(n+1)}2 \right)^2
,
[[/math]]
prove that
[[math]]
\sum_{i=1}^n i^3 = \left( \sum_{i=1}^n i \right)^2
.
[[/math]]
Verify this result directly for [math]n=1,2,3[/math].