Final answer:
We can prove that sn < k 1 for all n ∈ ℕ using the given hint: 9 10 9 102 ··· 9 10n = 1 - 1 10n for all n. We can rewrite sn as a series of terms, and by substituting the given hint equation, we can simplify and combine the terms to prove that sn < k 1. Therefore, sn < k 1 for all n ∈ ℕ.
Step-by-step explanation:
We can prove that sn < k 1 for all n ∈ ℕ using the given hint:
9 109 102 ··· 9 10n = 1 - 1/10n for all n
To prove sn < k 1, we need to show that sn - k 1 < 0. Using the given hint, we can rewrite sn as:
sn = k + 1/10 + 2/100 + 3/1000 + ... + n/10^n
Substituting the hint equation, we have:
sn = k + (1 - 1/10^n) + (2 - 2/10^n)/100 + (3 - 3/10^n)/1000 + ... + (n - n/10^n)/10^n
Simplifying, we get:
sn = k + 1 - 1/10^n + 2/100 - 2/10^n)/100 + 3/1000 - 3/10^n)/1000 + ... + (n - n/10^n)/10^n
Combining like terms, we have:
sn = (k + 1 + 2/100 + 3/1000 + ... + n/10^n) - (1/10^n + 2/10^n + 3/10^n + ... + n/10^n)
sn = (k + 1 + 2/100 + 3/1000 + ... + n/10^n) - (1 + 2 + 3 + ... + n)/10^n
sn = (k + 1 + 2/100 + 3/1000 + ... + n/10^n) - (n(n+1)/2)/10^n
sn = (k + 1 + 2/100 + 3/1000 + ... + n/10^n) - (n^2 + n)/2 * (1/10^n)
Now, we can see that the first term, (k + 1 + 2/100 + 3/1000 + ... + n/10^n), is clearly less than k 1 since it only contains positive terms. The second term, (n^2 + n)/2 * (1/10^n), is positive but decreases as n increases, therefore it is always less than k 1.
Therefore, sn < k 1 for all n ∈ ℕ.