How do you find the nth partial sum, determine whether the series converges and find the sum when it exists given #ln(4/3)+ln(9/8)+ln(16/15)+...+ln(n^2/(n^2-1))+...#?

Answer 1

To find the nth partial sum of the series ( \ln\left(\frac{n^2}{n^2-1}\right) ), we first express the general term ( a_n ) as ( \ln\left(\frac{n^2}{n^2-1}\right) ). Then, we find the nth partial sum by summing up the first ( n ) terms of the series.

The nth partial sum ( S_n ) is given by:

[ S_n = \ln\left(\frac{2^2}{2^2-1}\right) + \ln\left(\frac{3^2}{3^2-1}\right) + \ldots + \ln\left(\frac{n^2}{n^2-1}\right) ]

We can simplify each term using properties of logarithms. The general term can be written as:

[ \ln\left(\frac{n^2}{n^2-1}\right) = \ln(n^2) - \ln(n^2-1) ]

Then, we can rewrite ( S_n ) as:

[ S_n = \ln(2^2) - \ln(2^2-1) + \ln(3^2) - \ln(3^2-1) + \ldots + \ln(n^2) - \ln(n^2-1) ]

[ S_n = \ln(2^2 \cdot 3^2 \cdot \ldots \cdot n^2) - \ln(1 \cdot 2 \cdot \ldots \cdot (n^2-1)) ]

[ S_n = \ln\left(\frac{n!^2}{(n^2-1)!}\right) ]

Next, we need to determine whether the series converges. Since the nth partial sum ( S_n ) can be expressed in terms of factorials, we can use the Ratio Test to check for convergence.

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ]

[ \lim_{n \to \infty} \left| \frac{\ln\left(\frac{(n+1)!^2}{((n+1)^2-1)!}\right)}{\ln\left(\frac{n!^2}{(n^2-1)!}\right)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\ln((n+1)!^2) - \ln(((n+1)^2-1)!)}{\ln(n!^2) - \ln((n^2-1)!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{2\ln((n+1)!) - \ln(((n+1)^2-1)!)}{2\ln(n!) - \ln((n^2-1)!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{2\ln(n+1) + 2\ln(n!) - \ln(((n+1)^2-1)!)}{2\ln(n) + 2\ln(n!) - \ln((n^2-1)!)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{2\ln(n+1) - \ln(((n+1)^2-1)!)}{2\ln(n) - \ln((n^2-1)!)} \right| ]

Now, we need to compute the limits. If the limit is less than 1, the series converges. If it's greater than 1 or it doesn't exist, the series diverges.

[ \lim_{n \to \infty} \left| \frac{2\ln(n+1) - \ln(((n+1)^2-1)!)}{2\ln(n) - \ln((n^2-1)!)} \right| = \lim_{n \to \infty} \left| \frac{2\ln(n+1)}{2\ln(n)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\ln((n+1)^2)}{\ln(n^2)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\ln(n^2 + 2n + 1)}{\ln(n^2)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\ln(n^2 (1 + \frac{2}{n} + \frac{1}{n^2}))}{\ln(n^2)} \right| ]

[ = \lim_{n \to \infty} \left| \frac{\ln(n^2) + \ln(1 + \frac{2}{n} + \frac{1}{n^2})}{\ln(n^2)} \right| ]

[ = \lim_{n \to \infty} \left| 1 + \frac{\ln(1 + \frac{2}{n} + \frac{1}{n^2})}{\ln(n^2)} \right| ]

As ( n \to \infty ), ( \frac{2}{n} + \frac{1}{n^2} ) goes to 0. So, ( \ln(1 + \frac{2}{n} + \frac{1}{n^2}) ) goes to 0.

Therefore,

[ \lim_{n \to \infty} \left| 1 + \frac{\ln(1 + \frac{2}{n} + \frac{1}{n^2})}{\ln(n^2)} \right| = 1 ]

Since the limit equals 1, the Ratio Test is inconclusive. Additional tests might be needed to determine convergence or divergence.

To find the sum when it exists, we can attempt to simplify the nth partial sum further and evaluate it as ( n ) approaches infinity. However, given the complexity of the expression involving factorials, a closed-form expression for the sum might be difficult to obtain.

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Answer 2
The series is #sum_{n=2}^infty ln(n^2/{n^2-1})# The #N#th partial sum by using the properties of the logarithm is #ln(prod_{n=2}^N n^2/{n^2-1})=ln({prod_{n=2}^N n^2}/{prod_{n=2}^N(n+1)(n-1)})=ln({(N!)^2/4}/{{(N+1)N!}/4{N!}/(4N)})=ln({4N}/{N+1})# Hence # sum ln(n^2/{n^2-1})=lim_{N rightarrow infty} ln({4N}/{N+1})=2ln(2) #
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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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