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Exercise 375Calculus › Single-variable calculus › SequencesL1Added 2 Nov 2026

Three limits by the squeeze theorem

#squeeze theorem#sums#n-th roots

Statement

Compute the limit of the following sequences:
1. An=∑k=1nnn2+kA_n=\displaystyle\sum_{k=1}^n\frac n{n^2+k},
2. Bn=(1+2n+3n)1/nB_n=\left(1+2^n+3^n\right)^{1/n},
3. Cn=(a1n+⋯+apn)1/nC_n=\left(a_1^n+\dots+a_p^n\right)^{1/n} for fixed positive numbers a1,…,apa_1,\dots,a_p.

Hint

Bound every term of a sum by the smallest and the largest one.

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Solution

1. Each term satisfies nn2+n≤nn2+k≤nn2+1\frac n{n^2+n}\le\frac n{n^2+k}\le\frac n{n^2+1}. Summing the nn terms:
n2n2+n≤An≤n2n2+1.\frac{n^2}{n^2+n}\le A_n\le\frac{n^2}{n^2+1}.
Both bounds tend to 11, so An→1A_n\to1.

2. 3n≤1+2n+3n≤3⋅3n3^n\le1+2^n+3^n\le3\cdot3^n, so taking nn-th roots, 3≤Bn≤3⋅31/n3\le B_n\le3\cdot3^{1/n}, and 31/n→13^{1/n}\to1. So Bn→3B_n\to3.

3. Let M=max⁡aiM=\max a_i. Then Mn≤a1n+⋯+apn≤p MnM^n\le a_1^n+\dots+a_p^n\le p\,M^n, so M≤Cn≤p1/nMM\le C_n\le p^{1/n}M and p1/n→1p^{1/n}\to1. Hence Cn→MC_n\to M.

Remark. The result in 3 says that a sum of powers behaves, to the nn-th root, like its largest term: the same idea underlies the ℓp\ell^p norms tending to the maximum norm as p→∞p\to\infty.