Determine whether the sum from n=1 to infinity of arctan(2n^2/(25n+1)) converges or diverges. If it converges, what is the sum?
To determine whether the specified series converges or diverges, we can apply the Comparison Test.
1. First, let's evaluate the limit of the arctangent term as n approaches infinity:
lim(n→∞) arctan(2n^2/(25n+1))
The highest order term in the numerator and denominator is n^2. We can ignore the lower-order terms as n approaches infinity.
lim(n→∞) arctan(2n^2/(25n+1))
= lim(n→∞) arctan(2n^2/25n)
= lim(n→∞) arctan(2n/25)
= arctan(∞) (since n/∞ = 0)
The limit of arctan as x approaches infinity is π/2 (or 90 degrees). Therefore, the arctan term has a constant upper bound of π/2.
2. Now, let's consider the general term of the given series,
T(n) = arctan(2n^2/(25n+1))
3. We can compare T(n) to another series whose convergence is known.
Let's consider the series S(n) = π/2 * (2n^2/(25n+1))
By comparing T(n) to S(n), we can determine if the original series converges or diverges.
T(n) ≤ S(n) for all positive n since arctan(2n^2/(25n+1)) ≤ π/2 for all positive n.
4. Now, let's determine the convergence behavior of the series S(n):
S(n) = π/2 * (2n^2/(25n+1))
= πn^2/(25n+1)
To check if the series S(n) converges or diverges, we can apply the Ratio Test:
lim(n→∞) (a(n+1)/a(n)) = lim(n→∞) ((π(n+1)^2)/(25(n+1)+1)) * ((25n+1)/(πn^2))
= lim(n→∞) (π(n+1)^2/(πn^2))
= lim(n→∞) (n^2 + 2n + 1)/(n^2)
= 1
Since the limit is equal to 1, the Ratio Test is inconclusive.
5. However, since the original series T(n) is bounded above by the series S(n) which is inconclusive, the original series also converges.
Therefore, the series from n=1 to infinity of arctan(2n^2/(25n+1)) converges. However, we cannot determine the exact sum of the series using the given information.
To determine whether the given series converges or diverges, we can use the limit comparison test. Let's analyze the series and apply the test step by step.
The given series is the sum from n=1 to infinity of arctan(2n^2/(25n+1)). Let us define a new series b_n such that b_n = 2n^2/(25n+1).
First, we need to check if b_n is a positive sequence. Since both 2n^2 and (25n+1) are positive for all n ≥ 1, the ratio 2n^2/(25n+1) is also positive. Therefore, b_n is a positive sequence.
Next, we need to calculate the limit as n approaches infinity of b_n. We can find this limit by dividing both the numerator and denominator by n:
lim(n->∞) (2n^2/(25n+1))
= lim(n->∞) (2n/(25+1/n))
= lim(n->∞) (2n/25)
= ∞/25
= ∞
Since the limit is positive infinity, we can conclude that the series b_n diverges.
Now, we will apply the limit comparison test. Let a_n be the given series arctan(2n^2/(25n+1)), and considering that b_n diverges, we will compare a_n to b_n.
Using the limit comparison test, we take the limit as n approaches infinity of (a_n / b_n). If this limit is positive and finite, then the two series have the same nature.
lim(n->∞) (arctan(2n^2/(25n+1)) / (2n^2/(25n+1)))
= lim(n->∞) (arctan(2n^2/(25n+1)) * (25n+1) / (2n^2))
Now, we simplify the expression by removing the limits in case they exist, as well as canceling common factors:
= 25 * lim(n->∞) (arctan(2n^2/(25n+1)) / (2n^2))
Since we have previously determined that the series b_n diverges (limit is infinity), and the limit of the above expression is a finite positive value, we can conclude that the given series a_n also diverges.
Hence, the sum from n=1 to infinity of arctan(2n^2/(25n+1)) diverges.