Question

Suppose that \(\sum\limits_{n = 1}^\infty {{a_n}} \left( {{a_n} = 0} \right)\)is known to be a convergent series. Prove that \(\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \) is a divergent series.

Short answer

The value of adding or subtracting a value at least \(1\) to the partial sums can never converge.

Step-by-step solution

Definitions of converges and diverges

Converges: If a series has a limit, and the limit exists, the series converges.

Divergent: If a series does not have a limit, or the limit is infinity, then the series is divergent.

Derive the limits to check series

If \(\sum\limits_{n = 1}^\infty {{a_n}} \)is a convergent series

Then \(\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \) is a divergent series

\(\sum\limits_{n = 1}^\infty {{a_n}} \) is convergent, so we know that \(\mathop {\lim }\limits_{x \to \infty } {a_n} = 0\).

This is because as we go far out in the series, the contribution from each individual term must get smaller and smaller. So, that the sequence of partial sums.

Therefore, the series itself may converge.

Since \(\mathop {\lim }\limits_{x \to \infty } {a_n} = 0\), we know that there is some \({n_0}\) such that \(m > {n_0}\) implies that \(\left| {{a_m}} \right| < 1\).

Applying the definition of sequence convergence with \(\varepsilon  = 1\)

Applying the definition of sequence convergence with \(\varepsilon = 1\), then there is no way that \(\mathop {\lim }\limits_{x \to \infty } \frac{1}{{{a_n}}} = 0\).

Here, the term of \(\frac{1}{{{a_n}}}\)are always bigger than \(1\) after a certain point.

Then, the series \(\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \) cannot be convergent, since if it were convergent, we would have \(\mathop {\lim }\limits_{x \to \infty } {a_n} = 0\).

Actually, after\({n_0}\)we are adding or subtracting a value of at least\(1\)to the partial sums of\(\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \), so its partial sums can converge.

Therefore, \(\sum\limits_{n = 1}^\infty {\frac{1}{{{a_n}}}} \)is a divergent series.