For complex numbers \(\alpha_1,\ldots,\alpha_k\), prove that \[ \limsup_{n\to\infty} \left|\sum_{j=1}^k\alpha_j^n\right|^{1/n} =\max_{1\le j\le k}|\alpha_j|. \]
Solution
1. Put\[ R=\max_{1\leq j\leq k}\abs{\alpha_j}. \]If \(R=0\), then the asserted identity holds.
If \(R=0\), then every \(\alpha_j=0\). Hence \[ \sum_{j=1}^k\alpha_j^n=0 \] for every \(n\geq1\), so both sides of the required identity are zero.
2. Assume henceforth that \(R>0\) and define\[ \beta_j=\frac{\alpha_j}{R} \qquad (1\leq j\leq k). \]Then\[ \max_j\abs{\beta_j}=1. \]
By the definition of \(R\), \[ \abs{\beta_j} = \frac{\abs{\alpha_j}}{R} \leq1 \] for every \(j\), and equality holds for at least one \(j\).
3. One has\[ \limsup_{n\to\infty} \abs{\sum_{j=1}^k\beta_j^n}^{1/n} \leq1. \]
By step 2, \[ \abs{\sum_{j=1}^k\beta_j^n} \leq \sum_{j=1}^k\abs{\beta_j}^n \leq k. \] Therefore \[ \abs{\sum_{j=1}^k\beta_j^n}^{1/n} \leq k^{1/n}, \] and \(k^{1/n}\to1\).
4. Let \(\zeta_1,\ldots,\zeta_m\) be the distinct values among the\(\beta_j\) satisfying \(\abs{\beta_j}=1\), and let \(c_r\) be the multiplicityof \(\zeta_r\). Define\[ S_n=\sum_{r=1}^m c_r\zeta_r^n. \]Then\[ \lim_{N\to\infty} \frac1N\sum_{n=1}^N\abs{S_n}^2 = \sum_{r=1}^m c_r^2 >0. \]
4.1. For \(r\neq s\),\[ \lim_{N\to\infty} \frac1N\sum_{n=1}^N (\zeta_r\overline{\zeta_s})^n =0. \]
Since the \(\zeta_r\) are distinct and have modulus one, \[ q=\zeta_r\overline{\zeta_s} \] satisfies \(\abs{q}=1\) and \(q\neq1\). Thus \[ \frac1N\sum_{n=1}^Nq^n = \frac{q(1-q^N)}{N(1-q)}, \] whose modulus is at most \[ \frac{2}{N\abs{1-q}} \longrightarrow0. \]
4.2. The asserted mean-square limit holds.
4.1. Hence the limit is\[ \sum_{r=1}^m c_r^2. \]This number is positive because step 2 ensures that at least one\(\beta_j\) has modulus one, so \(m\geq1\).
4.3. Q.E.D.
Step 2 proves the claim of step 4.
5. There is a constant \(\delta>0\) such that\[ \abs{S_n}\geq2\delta \]for infinitely many positive integers \(n\).
Set \[ A=\sum_{r=1}^m c_r^2>0. \] By step 4, \[ \frac1N\sum_{n=1}^N\abs{S_n}^2\longrightarrow A. \] If \(\abs{S_n}<\sqrt{A/2}\) for all sufficiently large \(n\), then the Cesàro means of \(\abs{S_n}^2\) would have limsup at most \(A/2\), a contradiction. Thus \[ \abs{S_n}\geq\sqrt{A/2} \] for infinitely many \(n\). Taking \[ \delta=\frac12\sqrt{A/2} \] gives the claim.
6. If\[ T_n = \sum_{\abs{\beta_j}<1}\beta_j^n, \]then \(T_n\to0\).
The sum is finite, and each summand tends to zero because its base has modulus strictly smaller than one. Hence their sum tends to zero.
7. For infinitely many \(n\),\[ \abs{\sum_{j=1}^k\beta_j^n}\geq\delta. \]
By grouping the terms of modulus one according to equal values, \[ \sum_{j=1}^k\beta_j^n=S_n+T_n. \] Step 5 gives infinitely many \(n\) with \(\abs{S_n}\geq2\delta\), while step 6 gives \(\abs{T_n}<\delta\) for all sufficiently large \(n\). Therefore infinitely many of the indices from step 5 also satisfy \[ \abs{S_n+T_n} \geq \abs{S_n}-\abs{T_n} \geq\delta. \]
8. One has\[ \limsup_{n\to\infty} \abs{\sum_{j=1}^k\beta_j^n}^{1/n} \geq1. \]
Along the infinite subsequence from step 7, \[ \abs{\sum_{j=1}^k\beta_j^n}^{1/n} \geq \delta^{1/n}. \] Since \(\delta^{1/n}\to1\), the limsup is at least \(1\).
9. Therefore\[ \limsup_{n\to\infty} \abs{\sum_{j=1}^k\beta_j^n}^{1/n} =1. \]
The upper bound is step 3 and the lower bound is step 8.
10. The required value is\[ \boxed{ \limsup_{n\to\infty} \abs{\sum_{j=1}^k\alpha_j^n}^{1/n} = \max_{1\leq j\leq k}\abs{\alpha_j} }. \]
For \(R>0\), step 2 gives \[ \sum_{j=1}^k\alpha_j^n = R^n\sum_{j=1}^k\beta_j^n. \] Hence \[ \abs{\sum_{j=1}^k\alpha_j^n}^{1/n} = R\abs{\sum_{j=1}^k\beta_j^n}^{1/n}. \] Taking limsups and using step 9 yields the stated equality. The case \(R=0\) was proved in step 1.
11. Q.E.D.
Step 10 is the desired identity.