Chapter 2: Problem 13
Show that the following sequences are convergent and find their limits. (i) \(b_{n}:=\frac{1+a+a^{2}+\cdots+a^{n}}{n+1}\), where \(a \in(-1,1)\), (ii) \(b_{n}:=\frac{2^{n+1}-1}{(n+1) 2^{n}}\) (iii) \(b_{n}:=\frac{1}{n}\left(\frac{2}{5}+\frac{5}{11}+\cdots+\frac{n^{2}+1}{2 n^{2}+3}\right)\) (iv) \(b_{n}:=\frac{1}{n} \sum_{k=1}^{n} \frac{(k+1)^{k}}{k^{k}}\)
Short Answer
Step by step solution
Express the Geometric Sum
Finding the Limit
Simplify the expression
Finding the Limit
Comparing to a Known Convergent Sequence
Simplifying the Limit
Computing the Limit of\(b_n\)
Rearranging\(b_n\)
Applying the Stolz-Cesà ro Theorem
Finding the Limit
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Geometric Series
To find the sum of this series, we use the formula \( S_n = \frac{a(1- r^n)}{1-r} \) where \( S_n \) represents the sum of the first \( n \) terms. For \( |r| < 1 \) (which is the case here), the series has a limit as \( n \) approaches infinity, which speeds up the calculation of a potentially infinite sum. This demonstrates an important principle for convergent sequences: understanding how to express and manipulate them in familiar terms, like those of the geometric series, can greatly simplify finding their limits.
Limit of a Sequence
Part (ii) of the exercise incorporates this concept when simplifying \( \frac{2^{n+1}-1}{(n+1)2^n} \) and observing that as \( n \) approaches infinity, the term \( \frac{1}{2^n} \) gets vanishingly small compared to the other terms. Hence, the limit of the sequence \( b_n \) is 0 since all its terms eventually get infinitely close to 0. Recognizing the dominance of some terms over others as \( n \) increases is a central technique in finding sequence limits.
Riemann Sum
In the context of sequences, a Riemann sum arises when you sum over a sequence of terms that represent areas of slices of a curve over an interval, as demonstrated in part (iii) of the exercise with \( \frac{1}{n}(\frac{2}{5}+\frac{5}{11}+\cdots+\frac{n^2+1}{2n^2+3}) \). As \( n \) gets larger, these terms represent finer and finer divisions, approximating the integral of the function over the interval. The recognition of a Riemann sum in the exercise provides an avenue to convert a potentially complicated limit problem into the evaluation of an integral, which can often be more straightforward.
Stolz-Cesà ro Theorem
The sequence in part (iv) of the exercise is approached with this theorem. By first rearranging the terms of the sequence into a sum and then applying Stolz-Cesà ro, we turn a challenging limit evaluation into a tractable problem. This theorem is particularly valuable when dealing with sequences of quotients where the numerator and the denominator are both approaching infinity, providing a streamlined path to the answer.
Remember that for the theorem to be applicable, certain conditions about the sequences involved must be met, making it a powerful yet somewhat specialized tool in the mathematician's toolkit.