diverge or converge

Question

please provide if it diverges or converges and please provide steps.

∞

∑ sin (6n²+5 /n⁵+5)

n=1

Answer 1

I'm assuming the sum in question is ∑ sin[(6n² + 5)/(n⁵ + 5)].

I prefer the limit comparison test for series such as this.

Note 

lim _n→∞ sin[(6n² + 5)/(n⁵ + 5)] / [(6n² + 5)/(n⁵ + 5)]

= lim _x→0 sin(x)/x

= 1.

Since the limit is a positive finite number, the given series and the series ∑ [(6n² + 5)/(n⁵ + 5)] either both converge or both diverge.

Next, note

lim _n→∞ [(6n² + 5)/(n⁵ + 5)] / (1/n³)

= lim _n→∞ [(6n⁵ + 5n³)/(n⁵ + 5)]

= 6.

Since the limit is a positive finite number, the series ∑ [(6n² + 5)/(n⁵ + 5)] and the series ∑ (1/n³) either both converge or both diverge. But of course ∑ (1/n³) converges. So the given series converges.

Now, we should also check that the sequences (1/n³) and [(6n² + 5)/(n⁵ + 5)] and sin [(6n² + 5)/(n⁵ + 5)] have only positive terms, since this is a condition of the limit comparison test. This is obvious in the first two cases for all n ≥ 1.

To see that sin [(6n² + 5)/(n⁵ + 5)] is positive for all n ≥ 2, put f(x) = (6x² + 5)/(x⁵ + 5). Then

f'(x) = [x(-18x⁵ - 25x³ + 60)] / (x⁵ + 5)².

For x ≥ 2, the sign of f' is clearly negative while the sign of f is clearly positive. Together with the fact that f(2) = 29/37 < π, it follows that 0 < f(x) < π for all x ≥ 2. Thus, for n ≥ 2, we must have sin[ f(n) ] > 0.

Answer 2

In order for series to converge, the individual parts of the sum must go converge to zero, as n goes to infinity. In general terms,

∑_n=1^∞ a_n

converges only when lim a_n = 0 as n→∞.

However, lim (sin((6x²+5)/(x²+5)) = sin(6) ≠0.

The sum diverges.

Calculating the sum when n=1000 I got -282.68

and n = 10000 the sum is down to - 2797.44

In general, a series diverges, if for any large number M, we find n such that sum is greater than M. The same is in the opposite direction, for any large negative M, we find n such that the sum is smaller than M.

My assumption is that a_n = (6n² +5)/(n²+5).

In case, if there are no parenthesis, a_n = 6n² +5/n² +5

we still have lim (sin(6n² +5/n² +5) ≈ sin(6n² +5) ≠ k π.

What I mean by the above row, the points (6n²+5) is arbitrary far from points where sin(x) is zero. For that reason, we must say that the sum diverges.

Answer 3

The easiest way to show this converges is show that it is a positive series bounded above by a series that converges

As all values in the series exist (sin is between 0 and 1, and n^5 + 5 > 0 for n >= 1, so the fraction exists.

Consider n >= 7 (If the sum converges for n >= 7, it converges for all n, as the sum for n = 1 to 6 will be between -6 and 6).

For n >= 7, 6n^2 + 5 = n^2(6 + 5/n^2) < n^2(6 + 1) = 7n^2 <= n^3

For n >= 7, n^5 + 5 > n^5.

Thus, (6n^2 + 5)/(n^5 + 5) < n^3/n^5 = 1/n^2.

For n >= 7, 0 <= 1/n^2 <= 1/49, and for x on (0, 1/49], 0 < sin x < x

Thus, the sequence sin(6n^2 + 5)/(n^5 + 5) is bounded below by 0 and above by 1/n^2 on [7, infinity)∑∑

The sum of 1/n^2 from n = 7 to infinity converges. (If you wish, you can say that the sum is less than the integral from 6 to infinity of 1/n^2 = 1/6)

Thus, ∑ sin(6n^2 + 5)/(n^5 + 5) from n = 7 to infinity converges, and therefore, ∑ sin(6n^2 + 5)/(n^5 + 5) from n = 1 to infinity converges.