If $\sum_{r=1}^n T_r = \frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64}$, then $\lim_{n \to \infty} \sum_{r=1}^n \left(\frac{1}{T_r}\right)$ is equal to:

Step 1: Find the general term $T_n$

We are given the sum of the first $n$ terms, $S_n = \sum_{r=1}^n T_r$. To find the $n$ -th term $T_n$, we can subtract the sum of the first $n-1$ terms, $S_{n-1}$, from $S_n$. This formula allows us to isolate the $n$ -th term.

Step 2: Substitute $S_n$ and $S_{n-1}$ and simplify

Substitute the given expression for $S_n$ and $S_{n-1}$ into the formula for $T_n$. After simplifying the terms in $S_{n-1}$, we get $S_{n-1} = \frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64}$. We can then factor out common terms.

Step 3: Simplify $T_n$ further

By factoring out the common terms $\frac{(2n-1)(2n+1)(2n+3)}{64}$, we are left with $(2n+5) - (2n-3)$ inside the brackets, which simplifies to $8$. This leads to a simplified expression for $T_n$.

Step 4: Find $\frac{1}{T_r}$ using partial fractions

We need to find the reciprocal of $T_r$. To sum this series, we use partial fraction decomposition. The expression can be written as a difference of two terms, which is a common technique for telescoping series. We can verify this by combining the terms on the right side.

Step 5: Calculate the sum $\sum_{r=1}^n \frac{1}{T_r}$

This is a telescoping series. When we write out the terms, most of them will cancel out. For example, the second part of the first term will cancel with the first part of the second term, and so on. Only the first part of the first term and the second part of the last term will remain.

Step 6: Evaluate the telescoping sum

After cancellation, the sum simplifies to $2 \left[ \frac{1}{1 \cdot 3} - \frac{1}{(2n+1)(2n+3)} \right]$. This is because all intermediate terms cancel out.

Step 7: Calculate the limit as $n \to \infty$

As $n$ approaches infinity, the term $\frac{1}{(2n+1)(2n+3)}$ approaches $0$. Therefore, the limit of the sum is $2 \times \frac{1}{3}$, which simplifies to $\frac{2}{3}$.

More from this chapter

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