All the numerators equal \(1\) and the denominators are powers of \(2\), beginning with \(2^0\) as the denominator of the first term. So the \(n\)-th term is \(c_n = \frac{1}{2^{n-1}}\).
Look at the sequence of partial sums:
\(s_1 = 1 = 2 - 1\)
\(s_2 = 1 + \frac{1}{2} = \frac{3}{2} = 2 – \frac{1}{2}\)
\(s_3 = 1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4} = 2 – \frac{1}{4}\)
\(s_4 = 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{15}{8} = 2 – \frac{1}{8}\)
So \(\left\{s_n\right\}\) is an increasing sequence with general term \(s_n = 2 – \frac{1}{2^{n-1}}\).
Notice that \(s_n<2\) for all \(n\). Then the sequence cannot have a limit greater than \(2\). To see why this is so, suppose that some number \(L>2\) is the limit 0f \(\left\{s_n\right\}\). Then the interval \((2,L+1)\) is an interval containing \(L\) which does not contain a tail of the sequence, because all terms of the sequence are smaller than \(2\). So the sequence cannot have a limit larger than \(2\).
Since each term of the sequence \(s_n=\left\{2 – \frac{1}{2^{n-1}}\right\}\)is less than 2, perhaps it has a limit less than \(2\). However, this cannot be the case.
Suppose the sequence has a limit \(L<2\).
Let \((a,b)\) be an interval containing \(L\), but choose \(b\) between \(L\) and \(2\).
Since \(\frac{1}{2^{n-1}}\) can be made as small as we please by choosing \(n\) large enough, choose \(n\) sufficiently large that \(\frac{1}{2^{n-1}}<2-b\).
Then it follows that \(b<2-\frac{1}{2^{n-1}}=s_n\).
So the interval \((a,b)\) does not contain \(s_n\) or any term of the sequence coming after \(s_n\) (since it is an increasing sequence).
So \((a,b)\) does not contain a tail of the sequence.
Thus no number smaller than \(2\) is a limit of the sequence.
Therefore, either \(2\) is the limit of the sequence, or the sequence has no limit.
To prove that \(2\) is the limit of the sequence of partial sums, let \(( a, b )\) denote an interval containing the number \(2\). If we can show that the interval must contain a tail of the sequence \(\left\{ 2 -\frac{1}{2^{n-1}} \right\}\), then 2 must be the limit of the sequence. Since the only condition we have placed on the interval ( a, b ) is that it contain 2, anything we conclude must be true for any interval containing 2, no matter how small.
The key lies in the fact that the number \(\frac{1}{2^{n-1}}\) may be made as small as we please just by picking \(n\) large enough. Thus, we pick \(n\) large enough so that \(\frac{1}{2^{n-1}} < 2 – a\). Then it follows that \(a < 2 -\frac{1}{2^{n-1}} < 2\)
.Thus the \(n\)-th term of the sequence of partial sums, and every term thereafter lies in the interval \(( a, b )\). Thus the interval contains a tail of the sequence. So \(2\) is the limit of the sequence.