Question

(5) Is the following argument true or false? Give reasons for your answer. \begin{equation} \sum_{n=2}^{\infty} \frac{1}{n(n-1)}. \end{equation}Since $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ we can write this series as \begin{equation} \sum_{n=2}^{\infty} \frac{1}{n-1} - \sum_{n=2}^{\infty} \frac{1}{n} \end{equation}The second series is the divergent harmonic series. The first series has $a_n = \frac{1}{n-1}$ and since $(n - 1) < n$ we have $\frac{1}{n-1} > \frac{1}{n}$ and so this series diverges by comparison with the divergent harmonic series $\sum_{n=2}^{\infty} \frac{1}{n}$. Since both series diverge, their difference also diverges.

          (5) Is the following argument true or false? Give reasons for your answer.
\begin{equation*}
\sum_{n=2}^{\infty} \frac{1}{n(n-1)}.
\end{equation*}Since $\frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}$ we
can write this series as
\begin{equation*}
\sum_{n=2}^{\infty} \frac{1}{n-1} - \sum_{n=2}^{\infty} \frac{1}{n}
\end{equation*}The second series is the divergent harmonic series. The first series has $a_n = \frac{1}{n-1}$ and since $(n - 1) < n$ we have $\frac{1}{n-1} > \frac{1}{n}$ and so this series diverges by
comparison with the divergent harmonic series $\sum_{n=2}^{\infty} \frac{1}{n}$. Since both series diverge,
their difference also diverges.

(5) Is the following argument true or false? Give reasons for your answer.

∑n=2^∞(1)/(n(n-1)).
Since (1)/(n(n-1)) = (1)/(n-1) - (1)/(n) we
can write this series as

∑n=2^∞(1)/(n-1) - ∑n=2^∞(1)/(n)
The second series is the divergent harmonic series. The first series has an = (1)/(n-1) and since (n - 1) < n we have (1)/(n-1) > (1)/(n) and so this series diverges by
comparison with the divergent harmonic series ∑n=2^∞(1)/(n). Since both series diverge,
their difference also diverges.

Added by George C.

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