Course Material 2.6 Some Important Limits
We may discuss some standard results stating the convergence of some sequences
which will be helpful in perusing Real Analysis.
1. If 0 ≤ 𝑟 < 1 lim 𝑟 𝑛 = 0
𝑛→∞
2. If −1 < 𝑟 < 1 then lim 𝑟 𝑛 = 0
𝑛→∞
1 𝑛
3. lim (1 + 𝑛) = 𝑒
𝑛→∞
4. If −1 < 𝑟 < 1 then lim 𝑛𝑟 𝑛 = 0
𝑛→∞
1
5. If 𝑎 > 0 then lim 𝑎𝑛 = 1
𝑛→∞
1
6. lim 𝑛𝑛 = 1
𝑛→∞
𝑎𝑛
7. lim = 0 for any 𝑎 ∈ ℝ
𝑛→∞ 𝑛!
Proof.
2. Since −1 < 𝑟 < 1 then we have 0 ≤ |𝑟| < 1 then by (1) lim |𝑟|𝑛 = 0
𝑛→∞
Therefore lim |𝑟 𝑛 | = 0
𝑛→∞
Hence given 𝜖 > 0 there exists a positive integer 𝑁 such that for all 𝑛 ≥ 𝑁
|𝑟 𝑛 | < 𝜖 this means that |𝑟 𝑛 − 0| < 𝜖, proving that lim 𝑟 𝑛 = 0
𝑛→∞
3. This limit defines the Mathematical constant 𝑒. We can prove that the sequence
1 𝑛
{𝑥𝑛 }, where 𝑥𝑛 = (1 + ) converges to a limit that lies between 2 and 3
𝑛
By the Binomial expansion, we have
1 𝑛 𝑛! 1
𝑥𝑛 = (1 + ) = 1 + ∑𝑛𝑘=1
𝑛 𝑘!(𝑛−𝑘)! 𝑛𝑘
𝑛
1 1 2 𝑘−1
=1+∑ (1 − ) (1 − ) … . (1 − )
𝑘! 𝑛 𝑛 𝑛
𝑘=1
We claim that {𝑥𝑛 } is increasing and bounded between 2 and 3.
1 1 2 𝑘−1
a. 𝑥𝑛 = 1 + ∑𝑛𝑘=1 𝑘! (1 − 𝑛) (1 − 𝑛) … . (1 − )
𝑛
𝑛
1 1 2 𝑘−1
<1+∑ (1 − ) (1 − ) … . (1 − )
𝑘! 𝑛+1 𝑛+1 𝑛+1
𝑘=1
< 𝑥𝑛+1
Hence {𝑥𝑛 } is increasing
We may discuss some standard results stating the convergence of some sequences
which will be helpful in perusing Real Analysis.
1. If 0 ≤ 𝑟 < 1 lim 𝑟 𝑛 = 0
𝑛→∞
2. If −1 < 𝑟 < 1 then lim 𝑟 𝑛 = 0
𝑛→∞
1 𝑛
3. lim (1 + 𝑛) = 𝑒
𝑛→∞
4. If −1 < 𝑟 < 1 then lim 𝑛𝑟 𝑛 = 0
𝑛→∞
1
5. If 𝑎 > 0 then lim 𝑎𝑛 = 1
𝑛→∞
1
6. lim 𝑛𝑛 = 1
𝑛→∞
𝑎𝑛
7. lim = 0 for any 𝑎 ∈ ℝ
𝑛→∞ 𝑛!
Proof.
2. Since −1 < 𝑟 < 1 then we have 0 ≤ |𝑟| < 1 then by (1) lim |𝑟|𝑛 = 0
𝑛→∞
Therefore lim |𝑟 𝑛 | = 0
𝑛→∞
Hence given 𝜖 > 0 there exists a positive integer 𝑁 such that for all 𝑛 ≥ 𝑁
|𝑟 𝑛 | < 𝜖 this means that |𝑟 𝑛 − 0| < 𝜖, proving that lim 𝑟 𝑛 = 0
𝑛→∞
3. This limit defines the Mathematical constant 𝑒. We can prove that the sequence
1 𝑛
{𝑥𝑛 }, where 𝑥𝑛 = (1 + ) converges to a limit that lies between 2 and 3
𝑛
By the Binomial expansion, we have
1 𝑛 𝑛! 1
𝑥𝑛 = (1 + ) = 1 + ∑𝑛𝑘=1
𝑛 𝑘!(𝑛−𝑘)! 𝑛𝑘
𝑛
1 1 2 𝑘−1
=1+∑ (1 − ) (1 − ) … . (1 − )
𝑘! 𝑛 𝑛 𝑛
𝑘=1
We claim that {𝑥𝑛 } is increasing and bounded between 2 and 3.
1 1 2 𝑘−1
a. 𝑥𝑛 = 1 + ∑𝑛𝑘=1 𝑘! (1 − 𝑛) (1 − 𝑛) … . (1 − )
𝑛
𝑛
1 1 2 𝑘−1
<1+∑ (1 − ) (1 − ) … . (1 − )
𝑘! 𝑛+1 𝑛+1 𝑛+1
𝑘=1
< 𝑥𝑛+1
Hence {𝑥𝑛 } is increasing