Title of Question

Answer 1

In order to show convergence by the integral test, we need to show that the integral from 1 to infinity of x/2^x converges to a finite result (A decreasing monotonic function where a_n = f(n)). (Seems like ratio test is a lot easier)

You integrate by parts picking x as u and 2^-x dx as dv:

-x2^-x/ln(2) |₁^∞ - 2^-x/(ln(2))²|₁^∞ (The infinity limits go to 0)

1/(2ln2) + 1/(2(ln2)²) = (ln2 + 1)/(2(ln2)²) = 1.76...

The series can be worked out to be exactly 2.

If you take the sum of n terms, S_n and subtract 1/2xS_n you get 2(1-1/2ⁿ - n/2ⁿ⁺¹) which has a limit of 2 as n goes to infinity.

When you do the subtraction you get 1/2S_n = 1/2 + 1/4 + 1/8 +...-n/2ⁿ⁺¹ and the first part is just a geometric series with partial sum: 1/2(1-(1/2)ⁿ⁾/(1-1/2) = 1 - 1/2ⁿ

Answer 2

Here "k" is a variable or differential in terms of calculus, and "n" is a fixed constant that is referenced to an upper limit for a "k".

Answer 3

The sequence shall converge if the common ratio, r is 0 < | r | < 1.

Answer 4

Why do you think "k" is a variable? I think this is just some constant, the same as "C", which is usually in use.

By the way, the integral you have tells you exactly what your differential is - dx

Answer 5

Let f(x) be the integrand.

f(-x) = [(-x)²sin(-x) + (-x)⁵] / [(-x)⁴ + 5(-x)² + 1] = -f(x). So, f(x) is an odd function.

For odd functions (which are symmetric with respect to the origin), ∫_{(from -a to a)}f(x)dx = 0.

So the value of the given definite integral is 0.

Answer 6

Anon,

u substitutions don't seem to work.

WolframAlpah gives this answer, but ran out of computing time to solve it exactly.

Can anyone solve this exactly?

Answer 7

Let's use two different Geometric Series.

Finite Series:

S_n = a₀ + a₀r + a₀r² + ... + a₀r^n-1

S_n = a₀ (1 - rⁿ) / (1 - r)

Infinite Series:

S = a₀ + a₀r + a₀r² + a₀r³ _{+ ...}

S = a₀ / (1 - r)

As long as | r | < 1, the anwer will converge..

We must use the infinite series to answer the question.

A: Choose a₀ = 1, r = 1/3

S = 1 + (1/3) + (1/3)² + (1/3)³ + (1/3)⁴ ...

S = a₀ / (1 - r) = 1 / (1-1/3) = 1 / (2/3) = 3/2

B: Choose a₀ = 1, r = 1/4

S = 1 + (1/4) + (1/4)² + (1/4)³ + (1/4)⁴ + ...

S = a₀ / (1 - r) = 1 / (1-1/4) = 1 / (3/4) = 4/3

C = A + B = 3/2 + 4/3 = 17/6 = 2 5/6

Assuming these series are varying, this is a possible solution.

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