determine whether the series converges of diverges
the sum from n=1 to infinity of
1/(the square root of (n^3+1))
I said that through the comparision test (comparing to 1/the square root of (n^3) the series converges
To determine whether the series converges or diverges, let's use the comparison test.
First, we compare the given series to the series 1/sqrt(n^3).
Let's calculate the limit of the ratio of the terms:
lim (n → ∞) [(1 / sqrt(n^3 + 1)) / (1 / sqrt(n^3))]
= lim (n → ∞) ((sqrt(n^3)) / (sqrt(n^3 + 1)))
= lim (n → ∞) ((n^(3/2)) / ((n^3 + 1)^(1/2)))
To evaluate this limit, we can simplify it using the squeeze theorem.
Since n^(3/2) / (n^3 + 1)^(1/2) > 0, we can ignore the absolute values.
Consider the numerator:
n^(3/2) = √(n^3) = √(n^3 + 1 - 1)
We can rewrite the denominator as:
(n^3 + 1)^(1/2) = √(n^3 + 1)
Now, combining the numerator and denominator, we have:
n^(3/2) / (n^3 + 1)^(1/2) = (√(n^3 + 1 - 1)) / √(n^3 + 1)
Using the inequality √(a - b) > √a - √b, we can say that:
(√(n^3 + 1 - 1)) / √(n^3 + 1) > (√(n^3 + 1) - √1) / √(n^3 + 1)
Simplifying further, we get:
(√(n^3 + 1) - √1) / √(n^3 + 1) = (√(n^3 + 1) - 1) / √(n^3 + 1)
= (√(n^3 + 1) / (√(n^3 + 1))) - (1 / √(n^3 + 1))
= 1 - (1 / √(n^3 + 1))
We now have an expression that approaches 1 as n approaches infinity.
Therefore, lim (n → ∞) [(1 / sqrt(n^3 + 1)) / (1 / sqrt(n^3))] = 1
Since the limit is nonzero and finite, both series converge or diverge together.
However, since the series 1/sqrt(n^3) converges, we can conclude that the given series, sum from n=1 to infinity of 1/sqrt(n^3 + 1), also converges.
Hence, the series converges.
To determine whether the series converges or diverges, you correctly used the comparison test. The comparison test states that if you can find another series that either converges or diverges, and the original series behaves in a similar manner, then you can conclude the same behavior for the original series.
In this case, you compared the given series to the series 1/√(n^3). Let's examine the comparison in more detail:
Consider the series 1/√(n^3). We know that the sum of this series, as n approaches infinity, is a converging series since it is a p-series with p = 3/2 (exponent larger than 1).
Now, let's compare this series to the given series: 1/√(n^3+1). Notice that for any positive value of n, n^3 + 1 is always greater than n^3. Therefore, we can say that 1/√(n^3+1) is always less than 1/√(n^3).
Since the given series is always smaller than a converging series (1/√(n^3)), we can conclude that the given series also converges.