Finding the Sum of the Series \( \frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \ldots \)

We are asked to find the sum of the first \(n\) terms of the given series: \( \frac{1}{2}, \frac{3}{4}, \frac{7}{8}, \frac{15}{16}, \ldots \)

Let's look at the pattern of the terms in this series:

• The first term is \(a_1 = \frac{1}{2}\).
• The second term is \(a_2 = \frac{3}{4}\).
• The third term is \(a_3 = \frac{7}{8}\).
• The fourth term is \(a_4 = \frac{15}{16}\).

We can observe that the denominator of the \(k\)-th term is \(2^k\). The numerator seems to be one less than the denominator. So, the \(k\)-th term of the series, \(a_k\), can be written as: \( a_k = \frac{2^k - 1}{2^k} \)

This general term \(a_k\) can be simplified:

\( a_k = \frac{2^k}{2^k} - \frac{1}{2^k} = 1 - \frac{1}{2^k} \)

To find the sum of the first \(n\) terms, denoted by \(S_n\), we need to sum the general term \(a_k\) from \(k=1\) to \(n\):

\( S_n = \sum_{k=1}^{n} a_k = \sum_{k=1}^{n} \left(1 - \frac{1}{2^k}\right) \)

We can split this summation into two parts:

\( S_n = \sum_{k=1}^{n} 1 - \sum_{k=1}^{n} \frac{1}{2^k} \)

The first part is the sum of \(1\) added \(n\) times, which is simply \(n\).

\( \sum_{k=1}^{n} 1 = 1 + 1 + \ldots + 1 \quad (\text{n times}) = n \)

The second part is the sum of a geometric series: \( \sum_{k=1}^{n} \frac{1}{2^k} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \ldots + \frac{1}{2^n} \). This is a geometric series with the first term \(A = \frac{1}{2}\) and the common ratio \(R = \frac{1}{2}\). The sum of the first \(n\) terms of a geometric series is given by the formula \( S_{\text{geom}, n} = A \frac{1 - R^n}{1 - R} \), provided \(R \neq 1\).

Using this formula for our series:

\( \sum_{k=1}^{n} \frac{1}{2^k} = \frac{1}{2} \cdot \frac{1 - (\frac{1}{2})^n}{1 - \frac{1}{2}} = \frac{1}{2} \cdot \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} \)

\( = 1 \cdot \left(1 - \frac{1}{2^n}\right) = 1 - \frac{1}{2^n} \)

Now substitute the results of the two summations back into the expression for \(S_n\):

\( S_n = n - \left(1 - \frac{1}{2^n}\right) \)

\( S_n = n - 1 + \frac{1}{2^n} \)

Recall that \( \frac{1}{2^n} \) can also be written as \( 2^{-n} \). So, the sum of the first \(n\) terms is:

\( S_n = n - 1 + 2^{-n} \)

We can rearrange this as \( S_n = 2^{-n} + n - 1 \).

Let's quickly check for small values of n:

• For \(n=1\): \(S_1 = \frac{1}{2}\). Formula gives \(2^{-1} + 1 - 1 = \frac{1}{2} + 0 = \frac{1}{2}\). Matches.
• For \(n=2\): \(S_2 = \frac{1}{2} + \frac{3}{4} = \frac{2+3}{4} = \frac{5}{4}\). Formula gives \(2^{-2} + 2 - 1 = \frac{1}{4} + 1 = \frac{1+4}{4} = \frac{5}{4}\). Matches.
• For \(n=3\): \(S_3 = \frac{5}{4} + \frac{7}{8} = \frac{10+7}{8} = \frac{17}{8}\). Formula gives \(2^{-3} + 3 - 1 = \frac{1}{8} + 2 = \frac{1+16}{8} = \frac{17}{8}\). Matches.

The formula \( 2^{-n} + n - 1 \) correctly represents the sum of the first \(n\) terms of the given series.

The final answer is \( 2^{-n} + n - 1 \).

Revision Table: Series Summation

Additional Information: Types of Series and Sums

Understanding different types of series is fundamental in sequences and series. Here are a few key types:

• Arithmetic Series: A series where the difference between consecutive terms is constant. The sum of an arithmetic series can be found using formulas related to the first term, last term, and number of terms.
• Geometric Series: As discussed, a series where the ratio between consecutive terms is constant. Their sums are calculated using specific formulas involving the first term, common ratio, and number of terms.
• Harmonic Series: The series formed by the reciprocals of the positive integers: \(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots\). This series diverges.
• Telescoping Series: A series where intermediate terms cancel out, leaving only the first and last terms. The \(k\)-th term can often be expressed as a difference, e.g., \(a_k = f(k) - f(k+1)\).

Many series problems involve recognizing patterns, writing the general term, and then using summation properties or standard series sum formulas (like geometric series) to find the sum of the first \(n\) terms. Breaking down complex terms into simpler parts (like we did \( \frac{2^k-1}{2^k} = 1 - \frac{1}{2^k} \)) is a common technique.