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Telescoping Series and Divergence of $sum_{n1}^{infty} lnleft(frac{n}{n 1}right)$

April 28, 2025Technology4054
Does the Series $sum_{n1}^{infty} lnleft(frac{n}{n 1}right)$ Converge

Does the Series $sum_{n1}^{infty} lnleft(frac{n}{n 1}right)$ Converge or Diverge?

For every integer $n geq 2$, the natural logarithm function satisfies:

$$lnleft(frac{n}{n 1}right) ln(n) - ln(n 1)$$

Summing this expression from $n 1$ to $m$, we get:

$$sum_{n1}^{m} lnleft(frac{n}{n 1}right) ln(1) - ln(m 1) quad (1)$$

The term $ln(1)$ simplifies to 0, and thus the sum reduces to:

$$-ln(m 1)$$

As $m to infty$, $$-ln(m 1) to -infty$$

This indicates that the series $$sum_{n1}^{infty} lnleft(frac{n}{n 1}right)$$ diverges to $-infty$.

Further Analysis: Comparison with Harmonic Series

While the series diverges, we can perform additional analysis to provide a deeper understanding. Consider the inequality for natural logarithms:

$$forall n in mathbb{N}, ; n geq ln(n 1)$$

This implies that:

$$sum_{n1}^{infty} frac{1}{ln(n 1)} geq sum_{n1}^{infty} frac{1}{n 1}$$

The series on the right is a harmonic series shifted by 1, which diverges to $infty$.

Therefore, by the comparison test, we conclude that:

$$sum_{n1}^{infty} frac{1}{ln(n 1)} to infty$$

Mathematical Software Verification

Mathematica provides an easy way to verify the divergence of this series. Running the sum through Mathematica with an upper limit of $infty$ results in:

```mathematica Sum[1./Log[1. (n 1 Net])); Sum::div: Sum does not converge. ```

Similarly, plotting the partial sums for $n$ from 1 to $10^5$ shows a non-converging series. The Mathematica results support that the series does not converge.

Here is the Mathematica code used to plot the partial sums:

```mathematica f[m_] Sum[1./Log[n 1 Net), {n 1 m 1}]; p Table[{m, f[m]}, {m 1 10^5 100}]; ListLinePlot[p, AxesLabel {"m", "Partial Sum"}] ```

The graph clearly shows the series diverges as the partial sums grow without bound.